Question

a girl throws a ball with initial velocity v at an inclination of 45 degree. the ball strikes the smooth vertical wall at a horizontal distance d from the girl and after rebounding returns to her hand. what is the coefficient of restitution between the ball and the wall?

Answer 1

Answer:

The answer will be gd/v^2 -gd

Explanation:

According to the problem the ball is thrown with the velocity of v making 45 degree

Now the ball moves d distance from the girl

Therefore the vertical component of ball is vsin45degree and the horizontal component of the ball is vcos45degree

Now at the highest point the horizontal component will be reversed.

Therefore at the highest point the horizontal component will be evcos45degree

and the vertical component will be zero

Now the time taken by the ball to reach at the highest point t1 =  d/vcos45degree

Now for rebounding the time taken is t2 = d/evcos45degree

Now we will consider the vertical components

therefore

V(final)  = V(initial) - gt1

=> 0 = vsin45degree - gt1

=>t1 =v sin45degree/g

V(final)  = V(initial) + gt2

=> t2= v sin45degree/g

t1 +t2 = 2 vsin45degree/g = d/vcos45degree + d/evcos45degree

          => e = gd/v^2 -gd

Answer 2

AnsWEr:-

=> e = gd / v^2 - gd

I think this is helpful for you !