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1D Motion

Problems for Motion with Constant Acceleration in One Dimension

One-dimensional Motion with Constant Acceleration

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Problem : A bouncy ball has the property that if it hits the ground with velocity v , it bounces back up with velocity - .8v . If such a ball is dropped from a height h above the ground, how high will it bounce?

When the ball is initially dropped, its motion is governed by the position function for objects in free fall:

(1) x 1(t) = - gt 2 + h
When it bounces back up, it behaves like the example of a bullet being fired upwards, and hence obeys the equation:
(2) x 2(t) = - gt 2 + v 0 t
where v 0 = - .8v , and v is the velocity at which the ball initially hits the ground. To solve this problem, we must first use (1) to determine the velocity at which the ball hits the ground. Then we can plug this value into (2) and solve for the maximum height reached by the ball after it has bounced.

From (1) we can find two important pieces of information. First of all, v 1(t) = x 1'(t) = - gt . Secondly, setting x 1(t) = 0 we discover that the ball hits the ground at time t f = . Plugging this value into the velocity equation, we find that the velocity of the ball as it hits the ground is v = v 1(t f) = - . Hence,

v 0 = - .8v = .8.

Now we can move onto equation (2). We won't plug in the value for v 0 just yet, though. First we need to figure out at what point after the ball has bounced it will reach its maximum height. The following observation proves valuable: the ball's velocity is zero when it reaches its maximum height. To use this fact, we need to find v 2(t) and set this equation equal to zero in order to solve for the appropriate time. v 2(t) = x 2'(t) = - gt + v 0 , so solving for v 2(t) = 0 yields t max = , the time at which the ball reaches its maximum height after the bounce. Plugging this time back into the equation for x 2(t) yields the actual height the ball attains at this time:

x 2(t max) =

Now, plugging in the value we found previously for v 0 , we get our final answer:

h max = x 2(t max) = .64h (Notice that .64 = (.8)2!)

There's actually a much easier way to solve this problem, but it requires using physics principles that don't get introduced as early as kinematics. However, even though it requires quite a bit of effort, it is still quite remarkable that we can solve this problem with only our knowledge of one-dimensional kinematics!

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