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?

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  1. it would be very kind enough if you answer this question for me

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  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

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