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?
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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