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