Question

A ball is thrown from a 10 m high cliff with a speed of 14 m/s....

A ball is thrown from a 10 m high cliff with a speed of 14 m/s. How fast will the ball be going when it hits the ground?

A baseball is travelling at 40 m/s toward a batter. The batter swings and makes contact with the ball for 0.001 s, sending the ball back the other direction at 50 m/s. What is the ball’s change in momentum? How much force did the bat apply to the ball?

Homework Answers

Answer #1

here,

a)

initial speed of ball , u = 14 m/s

height of cliff , h0 = 10 m

let the speed of ball at the bottom of cliff be v

using third equation of motion

v^2 - u^2 = 2 * g * h0

v^2 - 14^2 = 2 * 9.81 * 10

solving for v

v= 19.8 m/s

the speed of ball when it hits the ground is 19.8 m/s

b)

initial speed , u = - 40 m/s

final speed , v = 50 m/s

the change in momentum , I = m * ( v - u)

I = m * ( 50 - (-40)) = 90 m kg.m/s

the force exerted , f = I/t

f = 9 * 10^4 m N

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