A ball is thrown vertically upwards from the top of a ledge with an initial velo

Question

A ball is thrown vertically upwards from the top of a ledge with an initial velocity of V A=35 ft/s Determine how high above the top of the cliff the ball will go before it stops at B. the time t AB it takes to reach its maximum height, and the total time t Ac needed for it to reach the ground at C from the instant is released.

Explanation / Answer

a) At top of the cliff point B , final velocity will be zero => Vf^2- Vi^2 = 2* acceleration*distance => Vf = 0 Distance = Vi^2 / 2*g = 35*35/(2*32.2) feet per second-squared = 1225/64.4 = 19.02 ft b) Vf -Vi = acceleration * time => time = 35/ 32.2 = 1.08 sec c) Total height = BC = 60 ft+ 19.02 ft = 79 ft Total height = Vi,B*t + (1/2)*acceleration*t^2 79 = 0 + (1/2)*32.2*t^2 t^2 = 79/(16.1) => t = 2.21 sec

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