**Problem : **
A bouncy ball has the property that if it hits the ground with velocity *v*,
it bounces back up with velocity - .8*v*. 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)When it bounces back up, it behaves like the example of a bullet being fired upwards, and hence obeys the equation:x_{1}(t) = -gt^{2}+h

(2)wherex_{2}(t) = -gt^{2}+v_{0}t

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,

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:

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!

Take a Study Break!