Two Ball Meeting


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.


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*t2+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*t2+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 at2 + 20 for a "released" ball where the initial velocity is zero. At a=-32 f/s/s, 20=16t2. t2=20/16=1.25 s2. 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=-16t2+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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