Projectile Motion

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Projectile Motion

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Explore projectile motion! Hit the target through learning about acceleration, trajectory, and velocity. Blast a car out of a cannon, and challenge yourself to hit a target! Learn about projectile motion by firing various objects. Set parameters such as angle, initial speed, and mass. Explore vector representations, and add air resistance to investigate the factors that influence drag.

Return to the Lessons Index Do the Lessons in Order Print-friendly page

Quadratic Word Problems: Projectile Motion (page 1 of 3)

Sections: Projectile motion, General word problems, Max/min problems

For our purposes, a 'projectile' is any object that is thrown, shot, or dropped. Usually the object is moving straight up or straight down.

Projectile Motion
  • An object is launched at 19.6 meters per second (m/s) from a 58.8-meter tall platform. The equation for the object's height s at time t seconds after launch is s(t) = –4.9t2 + 19.6t + 58.8, where s is in meters. When does the object strike the ground?

    • What is the height (above ground level) when the object smacks into the ground? Well, zero, obviously. Animal crossing. So I'm looking for the time when the height is s = 0. I'll set s equal to zero, and solve:

      0 = –4.9t2 + 19.6t + 58.8
      0 = t2 – 4t – 12
      0 = (t – 6)(t + 2)

    Then t = 6 or t = –2. The second solution is from two seconds before launch, which doesn't make sense in this context. (It makes sense on the graph, because the line crosses the x-axis at –2, but negative time won't work in this word problem.) So 't = –2' is an extraneous solution, and I'll ignore it.

      The object strikes the ground six seconds after launch.

Note the construction of the height equation in the problem above. The initial launch height was 58.8 meters, and the constant term was '58.8'. The initial velocity (launch speed) was 19.6 m/s, and the coefficient on the linear term was '19.6'. This is always true for these up/down projectile motion problems. (If you have an exercise with sideways motion, the equation will have a different form, but they'll always give you that equation.) The initial velocity is the coefficient for the middle term, and the initial height is the constant term. And the coefficient on the leading term comes from the force of gravity. This coefficient is negative, since gravity pulls downward, and the value will either be '4.9' (if your units are 'meters') or '16' (if your units are 'feet'). In general, the format is:

s(t) = –gt2 + v0t + h0

..where 'g' here is the '4.9' or the '16' derived from the value of the force of gravity (technically, it's half of the force of gravity, but you probably don't need to know that right now), 'v0' ('vee-naught', or 'vee-sub-zero') is the initial velocity, and 'h0' ('aitch-naught', or 'aitch-sub-zero') is the initial height.

Memorize this equation (or at least its meaning), because you may need to know this on the test.

  • An object in launched directly upward at 64 feet per second (ft/s) from a platform 80 feet high. What will be the object's maximum height? When will it attain this height?

    • Hmm.. They didn't give me the equation this time. But that's okay, because I can create the equation from the information that they did give me. The initial height is 80 feet above ground and the initial speed is 64 ft/s. Since my units are 'feet', then the number for gravity will be 16, and my equation is:

      s(t) = –16t2 + 64t + 80

    They want me to find the maximum height. For a negative quadratic like this, the maximum will be at the vertex of the upside-down parabola. So they really want me to find the vertex. From graphing, I know how to find the vertex; in this case, the vertex is at (2, 144):

      h = –b/2a = –(64)/2(–16) = –64/–32 = 2
      k = s(2) = –16(2)2 + 64(2) + 80 = –16(4) + 128 + 80 = 208 – 64 = 144

    But what does this vertex tell me? According to my equation, I'm plugging in time values and extracting height values, so the input '2' must be the time and the output '144' must be the height. Copyright © Elizabeth Stapel 2004-2011 All Rights Reserved

      Metal slug defense apk. It takes two seconds to reach the maximum height of 144 feet.

  • An object is launched from ground level directly upward at 39.2 m/s. For how long is the object at or above a height of 34.3 meters?

    My units this time are 'meters', so the gravity number will be '4.9'. Since the object started at ground level, the initial height was 0. Then my equation is:

      s(t) = –4.9t2 + 39.2t

    Advertisement

    Since this is a negative quadratic, the graph is an upside-down parabola. I can find the two times when the object is exactly 34.3 meters high, and I know that the object will be above 34.3 meters the whole time in between. Why 'two time', and how do I know that the time period is between those two times? Because the first time will be when the object passes a height of 34.3 meters on its way up to its maximum height, and the second time when be when it passes 34.3 meters as it is falling back down to the ground. So I have to solve the following:

      –4.9t2 + 39.2t = 34.3
      t2 – 8t + 7 = 0
      (t – 7)(t – 1) = 0

    Then the object is at 34.3 meters at one second after launch (going up) and againt at seven seconds after launch (coming back down). Subtracting to find the difference, I find that:

      The object is at or above 34.3 meters for six seconds.

Projectile

Don't be surprised if many of your exercises work out as 'neatly' as the above examples have. Many textbooks still engineer their exercises carefully, so that you can solve by factoring (that is, by quickly doing the algebra). However, heavy dependence on calculators is leading more texts to create 'interesting' (that is, needlessly complicated) exercises, so some (or all) of your exercises may involve much more messy computations than have been displayed here. If so, study these 'neat' examples carefully, until you are quite sure you follow the reasoning.

  • After the semester is over, you discover that the math department has changed textbooks (again) so the bookstore won't buy back your nearly-new book. You and your friend Herman decide to get creative. You go to the roof of a twelve-story building and look over the edge to the reflecting pool 160 feet below. You drop your book over the edge at the same instant that Herman chucks his book straight down at 48 feet per second. By how many seconds does his book beat yours into the water?

    • Our initial launch heights will be the same: we're both launching from 160 feet above ground. And the gravity number, since we're working in feet, will be 16. My initial velocity is zero, since I just dropped my book, but my buddy Herman's velocity is a negative48, the negative coming from the fact that he chucked his book down rather than up. So our 'height' equations are:

      mine: s(t) = –16t2 + 160
      his:       s(t) = –16t2 – 48t + 160

    In each case, I need to find the time for the books to reach a height of zero ('zero' being 'ground level'), so:

      mine: 0 = –16t2 + 160, t2 – 10 = 0, so t = ± sqrt(10)
      his:       0 = –16t2 – 48t + 160, t2 + 3t – 10 = 0, (t + 5)(t – 2) = 0, so t = –5 or t = 2

    I will ignore the negative time values. His book hits the water after two seconds, and mine hits after sqrt(10) seconds, or after about 3.16 seconds. That is:

      Herman's book hits the water about 1.16 seconds sooner than mine does.

Every once in a while, they'll get clever and put a 'projectile' problem into a different environment. The equation will remain the same in structure, but you may have to account for a different value for gravity.

  • The International Space Agency has finally landed a robotic explorer on an extra-solar planet. Some probes are extended from the lander's body to conduct various tests. To demonstrate the crushing weight of gravity on this planet, the lander's camera is aimed at a probe's ground-level ejection port, and the port launches a baseball directly upwards at 147 feet per second (ft/s), about the top speed of a professional pitcher. The force due to gravity on this planet is 98 ft/s2. Assuming no winds and that the probe can scurry out of the way in time, how long will it take for the ball to smack back into the surface?

    • To set up my equation for this exercise, I need to keep in mind that the value of the coefficient 'g' from the 'projectile motion' equation above is one-half of the value of the force due to gravity. In physics, there is the 'universal gravitational constant' G; then every object exerts its own gravitational force, which is related to its own mass and the universal constant G. In the 'projectile motion' formula, the 'g' is half of the value of the gravitational force for that particular body. For instance, the gravitational force on Earth is a downward 32 ft/s2, but we used '16' in the equation.

    So 'g' for my equation this time will by 98 ÷ 2 = 49. Then:

      s = –49t2 + 147t
      0 = –49t2 + 147t
      0 = t2 – 3t = t(t – 3)

    Then t = 0 or t = 3. The first solution represents when the ball was launched, so the second solution is the one I want.

      It takes three seconds for the ball to hit the ground.

Note: On Earth, it would take a little over nine seconds for the ball to fall back to the ground.

Top 1 2 3 Return to IndexNext >>

Cite this article as:

Stapel, Elizabeth. 'Quadratic Word Problems: Projectile Motion.' Purplemath. Available from
https://www.purplemath.com/modules/quadprob.htm. Accessed [Date] [Month] 2016

Popular Posts

  • Explore projectile motion! Hit the target through learning about acceleration, trajectory, and velocity. Blast a car out of a cannon, and challenge yourself to hit a target! Learn about projectile motion by firing various objects. Set parameters such as angle, initial speed, and mass. Explore vector representations, and add air resistance to investigate the factors that influence drag.

    Return to the Lessons Index Do the Lessons in Order Print-friendly page

    Quadratic Word Problems: Projectile Motion (page 1 of 3)

    Sections: Projectile motion, General word problems, Max/min problems

    For our purposes, a \'projectile\' is any object that is thrown, shot, or dropped. Usually the object is moving straight up or straight down.

    \'Projectile

    Note the construction of the height equation in the problem above. The initial launch height was 58.8 meters, and the constant term was \'58.8'. The initial velocity (launch speed) was 19.6 m/s, and the coefficient on the linear term was \'19.6'. This is always true for these up/down projectile motion problems. (If you have an exercise with sideways motion, the equation will have a different form, but they\'ll always give you that equation.) The initial velocity is the coefficient for the middle term, and the initial height is the constant term. And the coefficient on the leading term comes from the force of gravity. This coefficient is negative, since gravity pulls downward, and the value will either be \'4.9' (if your units are \'meters\') or \'16' (if your units are \'feet\'). In general, the format is:

    s(t) = –gt2 + v0t + h0

    ..where \'g' here is the \'4.9' or the \'16' derived from the value of the force of gravity (technically, it\'s half of the force of gravity, but you probably don\'t need to know that right now), \'v0' (\'vee-naught\', or \'vee-sub-zero\') is the initial velocity, and \'h0' (\'aitch-naught\', or \'aitch-sub-zero\') is the initial height.

    Memorize this equation (or at least its meaning), because you may need to know this on the test.

    • An object in launched directly upward at 64 feet per second (ft/s) from a platform 80 feet high. What will be the object\'s maximum height? When will it attain this height?

      • Hmm.. They didn\'t give me the equation this time. But that\'s okay, because I can create the equation from the information that they did give me. The initial height is 80 feet above ground and the initial speed is 64 ft/s. Since my units are \'feet\', then the number for gravity will be 16, and my equation is:

        s(t) = –16t2 + 64t + 80

      They want me to find the maximum height. For a negative quadratic like this, the maximum will be at the vertex of the upside-down parabola. So they really want me to find the vertex. From graphing, I know how to find the vertex; in this case, the vertex is at (2, 144):

        h = –b/2a = –(64)/2(–16) = –64/–32 = 2
        k = s(2) = –16(2)2 + 64(2) + 80 = –16(4) + 128 + 80 = 208 – 64 = 144

      But what does this vertex tell me? According to my equation, I\'m plugging in time values and extracting height values, so the input \'2' must be the time and the output \'144' must be the height. Copyright © Elizabeth Stapel 2004-2011 All Rights Reserved

        Metal slug defense apk. It takes two seconds to reach the maximum height of 144 feet.

    • An object is launched from ground level directly upward at 39.2 m/s. For how long is the object at or above a height of 34.3 meters?

      My units this time are \'meters\', so the gravity number will be \'4.9'. Since the object started at ground level, the initial height was 0. Then my equation is:

        s(t) = –4.9t2 + 39.2t

      Advertisement

      Since this is a negative quadratic, the graph is an upside-down parabola. I can find the two times when the object is exactly 34.3 meters high, and I know that the object will be above 34.3 meters the whole time in between. Why \'two time\', and how do I know that the time period is between those two times? Because the first time will be when the object passes a height of 34.3 meters on its way up to its maximum height, and the second time when be when it passes 34.3 meters as it is falling back down to the ground. So I have to solve the following:

        –4.9t2 + 39.2t = 34.3
        t2 – 8t + 7 = 0
        (t – 7)(t – 1) = 0

      Then the object is at 34.3 meters at one second after launch (going up) and againt at seven seconds after launch (coming back down). Subtracting to find the difference, I find that:

        The object is at or above 34.3 meters for six seconds.

    \'Projectile\'

    Don\'t be surprised if many of your exercises work out as \'neatly\' as the above examples have. Many textbooks still engineer their exercises carefully, so that you can solve by factoring (that is, by quickly doing the algebra). However, heavy dependence on calculators is leading more texts to create \'interesting\' (that is, needlessly complicated) exercises, so some (or all) of your exercises may involve much more messy computations than have been displayed here. If so, study these \'neat\' examples carefully, until you are quite sure you follow the reasoning.

    • After the semester is over, you discover that the math department has changed textbooks (again) so the bookstore won\'t buy back your nearly-new book. You and your friend Herman decide to get creative. You go to the roof of a twelve-story building and look over the edge to the reflecting pool 160 feet below. You drop your book over the edge at the same instant that Herman chucks his book straight down at 48 feet per second. By how many seconds does his book beat yours into the water?

      • Our initial launch heights will be the same: we\'re both launching from 160 feet above ground. And the gravity number, since we\'re working in feet, will be 16. My initial velocity is zero, since I just dropped my book, but my buddy Herman\'s velocity is a negative48, the negative coming from the fact that he chucked his book down rather than up. So our \'height\' equations are:

        mine: s(t) = –16t2 + 160
        his:         s(t) = –16t2 – 48t + 160

      In each case, I need to find the time for the books to reach a height of zero (\'zero\' being \'ground level\'), so:

        mine: 0 = –16t2 + 160, t2 – 10 = 0, so t = ± sqrt(10)
        his:         0 = –16t2 – 48t + 160, t2 + 3t – 10 = 0, (t + 5)(t – 2) = 0, so t = –5 or t = 2

      I will ignore the negative time values. His book hits the water after two seconds, and mine hits after sqrt(10) seconds, or after about 3.16 seconds. That is:

        Herman\'s book hits the water about 1.16 seconds sooner than mine does.

    Every once in a while, they\'ll get clever and put a \'projectile\' problem into a different environment. The equation will remain the same in structure, but you may have to account for a different value for gravity.

    • The International Space Agency has finally landed a robotic explorer on an extra-solar planet. Some probes are extended from the lander\'s body to conduct various tests. To demonstrate the crushing weight of gravity on this planet, the lander\'s camera is aimed at a probe\'s ground-level ejection port, and the port launches a baseball directly upwards at 147 feet per second (ft/s), about the top speed of a professional pitcher. The force due to gravity on this planet is 98 ft/s2. Assuming no winds and that the probe can scurry out of the way in time, how long will it take for the ball to smack back into the surface?

      • To set up my equation for this exercise, I need to keep in mind that the value of the coefficient \'g' from the \'projectile motion\' equation above is one-half of the value of the force due to gravity. In physics, there is the \'universal gravitational constant\' G; then every object exerts its own gravitational force, which is related to its own mass and the universal constant G. In the \'projectile motion\' formula, the \'g' is half of the value of the gravitational force for that particular body. For instance, the gravitational force on Earth is a downward 32 ft/s2, but we used \'16' in the equation.

      So \'g' for my equation this time will by 98 ÷ 2 = 49. Then:

        s = –49t2 + 147t
        0 = –49t2 + 147t
        0 = t2 – 3t = t(t – 3)

      Then t = 0 or t = 3. The first solution represents when the ball was launched, so the second solution is the one I want.

        It takes three seconds for the ball to hit the ground.

    Note: On Earth, it would take a little over nine seconds for the ball to fall back to the ground.

    Top 1 2 3 Return to IndexNext >>

    Cite this article as:

    Stapel, Elizabeth. \'Quadratic Word Problems: Projectile Motion.\' Purplemath. Available from
    https://www.purplemath.com/modules/quadprob.htm. Accessed [Date] [Month] 2016

    ...'>Projectile Motion(16.03.2020)

  • Explore projectile motion! Hit the target through learning about acceleration, trajectory, and velocity. Blast a car out of a cannon, and challenge yourself to hit a target! Learn about projectile motion by firing various objects. Set parameters such as angle, initial speed, and mass. Explore vector representations, and add air resistance to investigate the factors that influence drag.

    Return to the Lessons Index Do the Lessons in Order Print-friendly page

    Quadratic Word Problems: Projectile Motion (page 1 of 3)

    Sections: Projectile motion, General word problems, Max/min problems

    For our purposes, a \'projectile\' is any object that is thrown, shot, or dropped. Usually the object is moving straight up or straight down.

    \'Projectile

    Note the construction of the height equation in the problem above. The initial launch height was 58.8 meters, and the constant term was \'58.8'. The initial velocity (launch speed) was 19.6 m/s, and the coefficient on the linear term was \'19.6'. This is always true for these up/down projectile motion problems. (If you have an exercise with sideways motion, the equation will have a different form, but they\'ll always give you that equation.) The initial velocity is the coefficient for the middle term, and the initial height is the constant term. And the coefficient on the leading term comes from the force of gravity. This coefficient is negative, since gravity pulls downward, and the value will either be \'4.9' (if your units are \'meters\') or \'16' (if your units are \'feet\'). In general, the format is:

    s(t) = –gt2 + v0t + h0

    ..where \'g' here is the \'4.9' or the \'16' derived from the value of the force of gravity (technically, it\'s half of the force of gravity, but you probably don\'t need to know that right now), \'v0' (\'vee-naught\', or \'vee-sub-zero\') is the initial velocity, and \'h0' (\'aitch-naught\', or \'aitch-sub-zero\') is the initial height.

    Memorize this equation (or at least its meaning), because you may need to know this on the test.

    • An object in launched directly upward at 64 feet per second (ft/s) from a platform 80 feet high. What will be the object\'s maximum height? When will it attain this height?

      • Hmm.. They didn\'t give me the equation this time. But that\'s okay, because I can create the equation from the information that they did give me. The initial height is 80 feet above ground and the initial speed is 64 ft/s. Since my units are \'feet\', then the number for gravity will be 16, and my equation is:

        s(t) = –16t2 + 64t + 80

      They want me to find the maximum height. For a negative quadratic like this, the maximum will be at the vertex of the upside-down parabola. So they really want me to find the vertex. From graphing, I know how to find the vertex; in this case, the vertex is at (2, 144):

        h = –b/2a = –(64)/2(–16) = –64/–32 = 2
        k = s(2) = –16(2)2 + 64(2) + 80 = –16(4) + 128 + 80 = 208 – 64 = 144

      But what does this vertex tell me? According to my equation, I\'m plugging in time values and extracting height values, so the input \'2' must be the time and the output \'144' must be the height. Copyright © Elizabeth Stapel 2004-2011 All Rights Reserved

        Metal slug defense apk. It takes two seconds to reach the maximum height of 144 feet.

    • An object is launched from ground level directly upward at 39.2 m/s. For how long is the object at or above a height of 34.3 meters?

      My units this time are \'meters\', so the gravity number will be \'4.9'. Since the object started at ground level, the initial height was 0. Then my equation is:

        s(t) = –4.9t2 + 39.2t

      Advertisement

      Since this is a negative quadratic, the graph is an upside-down parabola. I can find the two times when the object is exactly 34.3 meters high, and I know that the object will be above 34.3 meters the whole time in between. Why \'two time\', and how do I know that the time period is between those two times? Because the first time will be when the object passes a height of 34.3 meters on its way up to its maximum height, and the second time when be when it passes 34.3 meters as it is falling back down to the ground. So I have to solve the following:

        –4.9t2 + 39.2t = 34.3
        t2 – 8t + 7 = 0
        (t – 7)(t – 1) = 0

      Then the object is at 34.3 meters at one second after launch (going up) and againt at seven seconds after launch (coming back down). Subtracting to find the difference, I find that:

        The object is at or above 34.3 meters for six seconds.

    \'Projectile\'

    Don\'t be surprised if many of your exercises work out as \'neatly\' as the above examples have. Many textbooks still engineer their exercises carefully, so that you can solve by factoring (that is, by quickly doing the algebra). However, heavy dependence on calculators is leading more texts to create \'interesting\' (that is, needlessly complicated) exercises, so some (or all) of your exercises may involve much more messy computations than have been displayed here. If so, study these \'neat\' examples carefully, until you are quite sure you follow the reasoning.

    • After the semester is over, you discover that the math department has changed textbooks (again) so the bookstore won\'t buy back your nearly-new book. You and your friend Herman decide to get creative. You go to the roof of a twelve-story building and look over the edge to the reflecting pool 160 feet below. You drop your book over the edge at the same instant that Herman chucks his book straight down at 48 feet per second. By how many seconds does his book beat yours into the water?

      • Our initial launch heights will be the same: we\'re both launching from 160 feet above ground. And the gravity number, since we\'re working in feet, will be 16. My initial velocity is zero, since I just dropped my book, but my buddy Herman\'s velocity is a negative48, the negative coming from the fact that he chucked his book down rather than up. So our \'height\' equations are:

        mine: s(t) = –16t2 + 160
        his:         s(t) = –16t2 – 48t + 160

      In each case, I need to find the time for the books to reach a height of zero (\'zero\' being \'ground level\'), so:

        mine: 0 = –16t2 + 160, t2 – 10 = 0, so t = ± sqrt(10)
        his:         0 = –16t2 – 48t + 160, t2 + 3t – 10 = 0, (t + 5)(t – 2) = 0, so t = –5 or t = 2

      I will ignore the negative time values. His book hits the water after two seconds, and mine hits after sqrt(10) seconds, or after about 3.16 seconds. That is:

        Herman\'s book hits the water about 1.16 seconds sooner than mine does.

    Every once in a while, they\'ll get clever and put a \'projectile\' problem into a different environment. The equation will remain the same in structure, but you may have to account for a different value for gravity.

    • The International Space Agency has finally landed a robotic explorer on an extra-solar planet. Some probes are extended from the lander\'s body to conduct various tests. To demonstrate the crushing weight of gravity on this planet, the lander\'s camera is aimed at a probe\'s ground-level ejection port, and the port launches a baseball directly upwards at 147 feet per second (ft/s), about the top speed of a professional pitcher. The force due to gravity on this planet is 98 ft/s2. Assuming no winds and that the probe can scurry out of the way in time, how long will it take for the ball to smack back into the surface?

      • To set up my equation for this exercise, I need to keep in mind that the value of the coefficient \'g' from the \'projectile motion\' equation above is one-half of the value of the force due to gravity. In physics, there is the \'universal gravitational constant\' G; then every object exerts its own gravitational force, which is related to its own mass and the universal constant G. In the \'projectile motion\' formula, the \'g' is half of the value of the gravitational force for that particular body. For instance, the gravitational force on Earth is a downward 32 ft/s2, but we used \'16' in the equation.

      So \'g' for my equation this time will by 98 ÷ 2 = 49. Then:

        s = –49t2 + 147t
        0 = –49t2 + 147t
        0 = t2 – 3t = t(t – 3)

      Then t = 0 or t = 3. The first solution represents when the ball was launched, so the second solution is the one I want.

        It takes three seconds for the ball to hit the ground.

    Note: On Earth, it would take a little over nine seconds for the ball to fall back to the ground.

    Top 1 2 3 Return to IndexNext >>

    Cite this article as:

    Stapel, Elizabeth. \'Quadratic Word Problems: Projectile Motion.\' Purplemath. Available from
    https://www.purplemath.com/modules/quadprob.htm. Accessed [Date] [Month] 2016

    ...'>Projectile Motion(16.03.2020)