# 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 .
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 $\Delta y$.