Question
a stone is dropped from a 75m building. when a stone has dropped 15m, a second stone was thrown downward with an initial velocity such that the two stones hit the ground at the same time. what was the initial velocity of the second stone.
Answers
d = Vo*t + 4.9t^2 = 15,
0 + 4.9t^2 = 15,
t^2 = 3.06,
t = 1.75s. = Time to fall 15m.
Therefore, the 1st stone had a 1.75s
headstart:
d1 = 0 + 4.9t^2 = 75
t^2 = 15.31,
t1 = 3.91s.
t2 = 3.91 - 1.75 = 2.16s.
d2 = Vo*t + 4.9t^2 = 75,
2.16*Vo + 4.9(2.16)^2 = 75,
2.16Vo + 22.9 = 75,
2.16*Vo = 75 - 22.9 = 52.1,
Vo = 24.1m/s. = Initial velocity of 2nd stone
0 + 4.9t^2 = 15,
t^2 = 3.06,
t = 1.75s. = Time to fall 15m.
Therefore, the 1st stone had a 1.75s
headstart:
d1 = 0 + 4.9t^2 = 75
t^2 = 15.31,
t1 = 3.91s.
t2 = 3.91 - 1.75 = 2.16s.
d2 = Vo*t + 4.9t^2 = 75,
2.16*Vo + 4.9(2.16)^2 = 75,
2.16Vo + 22.9 = 75,
2.16*Vo = 75 - 22.9 = 52.1,
Vo = 24.1m/s. = Initial velocity of 2nd stone
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