Question

Suppose you throw a stone straight up with an initial speed of 15 m s−1
.If you throw the second stone 1.30 s after the first, with what speed must you throw the second
stone if it is to hit the first at a height of 11.0 m?

Answer 1

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Answer 2

the 1st stone reaches its peak approx. 1.5 s after being thrown
... so the 2nd stone hits the 1st during the 1st stone's descent

time(s) for 1st stone to reach 11.0 m ... 11.0 = (1/2 * g * t^2) + 15 t
... remember, g is acceleration downward (negative)
... find the downward (larger) time
... subtract 1.30 s to find the upward flight time of the 2nd stone (Ts)

2nd stone ... 11.0 = [1/2 * g * (Ts)^2] + (v * Ts)
... solve for the launch velocity (v) of the second stone