Asked by Abdullah
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?
.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?
Answers
Answered by
Abdullah
it would be very kind enough if you answer this question for me
Answered by
R_scott
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
... 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
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.