Both travel 11m,
11=15t-1/2 g t^2 so solve for the time the first stone is at 11 meters, You should get two times, going up, and coming down.
Now, for the second stone
1) hitting the first stone going up..
11=v*time-1/2 g time^2 USING THE SHORTER of the two times found, solve for v.
2) Hitting the first stone on the way down...use the longer of the times.
Suppose you throw a stone straight up with an initial speed of 15.00 m/s
If you throw a second stone straight up 1.00s 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?
2 answers
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