Question

A rock is thrown up from the edge of a tall building. the rock reaches its maximum height 1.os after being thrown. then, when it goes down, the rock hits the ground 9.0s after being thrown. how fast is the rock thrown? what would be the maximum height above the build, which would the rock reach? what is the height of the building?

Answer 1

a)

v = gt = 9.8* 1 = 9.8 m/s

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b)

max height above building

h = v^2/2g = 9.8^2 / ( 2* 9.8)

h = 4.9 m

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c)

using 2nd equation of motion

0 = H + vt - 0.5 gt^2

0 = H + 9.8* 9 - 4.9* 9^2

H = 308.7 m

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