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A girl throws a ball an initial velocity v at an angle of 4 5 ∘ . The ball strikes a smooth vertical wall at a horizontal distance d from the girl and after rebound returns to her hands. What is the coefficient of restitution between wall and ball? ( v 2 > d g )

A girl throws a ball an initial velocity v at an angle of 45. The ball strikes a smooth vertical wall at a horizontal distance d from the girl and after rebound returns to her hands. What is the coefficient of restitution between wall and ball? (v2>dg)

Grade:11

1 Answers

Heer Shah
19 Points
one month ago
Let the wall be d distance away.
When ball will strike the wall its horizontal velocity will be reversed and will become ev (where v is the velocity of ball just before collision) and vertical velocity remains same.
Vx=vcos45
Before collision
Vx=d/t1...........(1)
After collision  
eVx=d/t2........(2)
 
Vy=vsin45
Now net displacement in y =0 so
0=vy(t1+t2)-0.5g(t1+t2)².......(3)
 
From eqn 1&2:  t1+t2 =d/Vx +d/eVx
From eqn 3:  t1+t2=2Vy/g
 
d/Vx (1+1/e)=2Vy/g
(1+1/e)=2×vsin45×vcos45/dg
1+1/e =v²/dg
1/e =(v²-dg)/dg
e=dg/(v²-dg)

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