Heer Shah
Last Activity: 3 Years 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)