A ball is thrown upward from the ground with an initial speed of 25m/s; at the s

Question

A ball is thrown upward from the ground with an initial speed of 25m/s; at the same instant, a ball is dropped from rest from a building 15m high. After how long will the balls be at the same height? I really just dont understand how to do this problem. Thanks for anyone who tries to help. A ball is thrown upward from the ground with an initial speed of 25m/s; at the same instant, a ball is dropped from rest from a building 15m high. After how long will the balls be at the same height? I really just dont understand how to do this problem. Thanks for anyone who tries to help.

Explanation / Answer

Let's use the equation x = xo + vot + 1/2at^2.

Ball A: x = 0 + 25t + 1/2 * -9.8 * t^2
Ball B: x = 15 + 0t + 1/2 * -9.8 * t^2

Since they both equal x, they can be set to equal each other.

-4.9t^2 + 25t = -4.9t^2 + 15
25t = 15
t = 0.6 s

They will be at the same height at 0.6 seconds.

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