**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:

Now, plugging in the value we found previously for

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!