Two Ball Meeting

## Question:

Ball A was dropped from a height of 40 feet. At the same time
Ball B was thrown upward from a height 5 feet above the ground.
The balls pass each other at a height of 20 feet. What was the
speed with which ball b was thrown upward.
## Answer:

Here are some facts we will need to use. In this problem we are
dealing with objects subject to gravity so we have a constant
acceleration(a) in the downward direction. Objects subject to
constant acceleration have a linear change in velocity. That
means that the average velocity may be applied to a whole event,
avoiding a lot of calculus. The average velocity over the time
interval (t2-t1) is (V2+V1)/2 so an object under constant
acceleration, between time 1 and time 2 will travel (V1+V2)/2*(t2-t1). The old "distance is
speed times time" thing. But V2 is
V1+a(t2-t1), so plugging that in to our expression for
distance traveled(d) we get
d=1/2*(V1+a*(t2-t1))*(t2-t1) or
d=1/2*a*(t2-t1)^{2}+V1*(t2-t1). If we take t1 to
be zero this cleans up to
d=1/2*a*t^{2}+V1*t. To calculate the position of
an object subject to constant acceleration, we need to add the
distance traveled to its initial position (x0) so x=1/2*a*t^{2}+v0*t+x0.
Now to the problem at hand. The meeting of the balls takes
place at the time ball A reaches the 20 foot altitude. Ball A
will take how long to fall twenty feet? x=1/2
at^{2} + 20 for a "released" ball where
the initial velocity is zero. At a=-32
f/s/s, 20=16t^{2}. t^{2}=20/16=1.25 s^{2}. t=1.12 seconds. Now how fast must I throw a
ball upward so that it reaches 20 feet in 1.12 seconds. Here
x=-16t^{2}+v0t+5. Plug in 1.12
seconds for t and solve for v0. I get
1.12v0=40 or v0=35.7 f/s.

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