A person, standing on a vertical cliff a height h above a lake, wants to jump into the lake but notices a rock just at the surface level with its furthest edge a distance s from the bottom of the cliff. The person realizes that with a running start it will be possible to just clear the rock, so the person steps back from the edge a distance d and starting from rest, runs at a constant acceleration a and then leaves the cliff horizontally. The person just clears the rock. Find s in terms of the given quantities d , a , h , and the gravitational acceleration g

asked by ss01

11 years ago

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1 answer

0.4.9*t^2 = h
t^2 = h/4.9
t = sqrt (h)/2.21 = Fall time.

S = Vx * t

answered by Henry

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Question

A person, standing on a vertical cliff a height h above a lake, wants to jump into the lake but notices a rock just at the surface level with its furthest edge a distance s from the bottom of the cliff. The person realizes that with a running start it will be possible to just clear the rock, so the person steps back from the edge a distance d and starting from rest, runs at a constant acceleration a and then leaves the cliff horizontally. The person just clears the rock. Find s in terms of the given quantities d , a , h , and the gravitational acceleration g

1 answer

0.4.9*t^2 = h
t^2 = h/4.9
t = sqrt (h)/2.21 = Fall time.

S = Vx * t

Henry · Answer

0.4.9*t^2 = h t^2 = h/4.9 t = sqrt (h)/2.21 = Fall time. S = Vx * t