Question

A ball rolls without slipping down a ramp with a 20 degree slope. If the ball starts 2.0m higher than the end of the ramp and the ramp is 3.0m above the ground, how far out from the end of the ramp does the ball land? I(ball)= 2/5 mr^2

Answer 1

first need to find ball speed at as it reach  bottom of ramp

for that apply energy conservation

energy at top = energy at bottom of ramp

mgh = 1/2*m*v^2 +1/2*I*ω^2 (  ω =v/r because ball rolls without slipping )

mgh = 1/2*m*v^2 + 1/2* 2/5*mr^2 * (v/r)^2

gh = 0.5v^2 +0.2v^2

from here  

v = √(gh/0.7) = √ ( 9.8*2 /0.7) =5.29 m/s at the bottom of ramp  

ball only have horizontal velocity , no vertical velocity at the bottom of ramp

Vx = 5.29 m/s

ramp is 3 m above of the ground so

by kinematics time taken by ball to reach the ground

sy =uy*t+ 1/2*g*t^2

3 = 0 +0.5*9.8*t^2

t = sqrt( 2*3/ 9.8) = 0.782460796 s

so distance traveled horizontaly by ball in this time  

= Vx*t = 5.29*0.782460796 =4.13921761 m answer