A boy standing at a distance d = 4 m from the bottom of the building throws a ba
ID: 1856903 • Letter: A
Question
A boy standing at a distance d = 4 m from the bottom of the building throws a ball against the wall of the building. He releases the ball with a velocity of 15 m/s at an angle of 30 degree with respect to the horizontal. If the coefficient of restitution between the ball and the building is e = 0.4. determine the velocity of the ball immediately after it bounces off the wall. Also, calculate the horizontal distance from the wall of the point where the ball strikes the ground. The boy initially releases the ball from a height of 1.5 m above the ground.Explanation / Answer
time to hit the wall
4 = 15*cos(30) t
t =0.308 s
hits at y= 1.5+15*sin(30)*0.308-0.5*9.81*.308^2=3.34
hits with velocity vx = 15*cos(30) = 12.99 m/s
vy = 15*sin(30) -9.81*0.308=4.48 m/s
v before = sqrt(12.99^2 +4.48^2)=13.74
so vafter = 0.4*13.74 = 5.50 m/s
vy doesnt change but vx does
sqrt(4.48^2 + v^2) = 5.5
v=-3.19 m/s
hits the ground when y = 0
0 = 0 + 4.48*t - 0.5*9.81*t^2
t=0.913 s
x = 0.913*3.19=2.91 m from wall
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