# A ball is thrown vertically upward

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A ball is thrown vertically upward with an initial speed of 23 m/s. Then, 1.6 s later, a stone is thrown straight up (from the same initial height as the ball) with an initial speed of 37 m/s .
How far above the release point will the ball and stone pass each other? The acceleration of gravity is 9.8 m/s2 .
Answer in units of m
asked Sep 11, 2014

## 1 Answer

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Set the time when the first ball is launched to be t=0. Set the time they meet to be t=T (assume they will meet).

If both balls meet at t=T, their vertical displacement would be the same.

Therefore $\Delta y_1=\Delta y_2$, which gives $$V_1T-\frac 12 gT^2=V_2(T-1.6)-\frac12 g(T-1.6)^2$$

Solve for T and then find $\Delta y$.
answered Sep 11, 2014 by (21,750 points)
I do not get it can you explain it more clearly