Question

A boy on a cliff 490 m high drops a stone. One second later, he throws a second stone after the first. They both hit the ground at the same time. With what speed did he throw the second stone ?
(g = 9.8 ms–2)

Answer 1

Speed at which the second stone was thrown = 12.101 m/s

Explanation:

First stone

s = ut + 1/2 gt^2

As u =0, we get 49 = 1/2*9.8*t^2

t^2 = 98 / 9.8 = 10

So time = root of 10 seconds

Second stone

t = (root of 10 - 1) seconds = 3.16 -1 = 2.16 seconds

49 = ux(2.16) + 1/2*9.8*(2.16)^2

49 = 2.16ux + 1/2*9.8x(2.16)^2

49 = 2.16ux + 22.8614

u = 12.101 m/s

Speed at which the second stone was thrown = 12.101 m/s