Question

A ball collides elastically with a massive wall moving towards it with a velocity of v as shown. The collision occurs at a height of h above ground level and the velocity of the ball just before collision is 2v in horizontal direction. Then the distance between the foot of the wall and the point on the ground where the ball lands (at the instant the ball lands) is $n\times v\sqrt{\frac{2h}{g}},$ where n is (Answer upto two digit after decimal places)

Answer 1

Solve in the reference frame fixed to the wall.
Before collision, velocity of ball = 3v towards it.
$\therefore$ After elastic collision of ball = 3v away from it (here we have used the result that a light mass is reversed by a heavy body at Time of flight $=\sqrt{\frac{2h}{g}}$
$\therefore$ Distance between wall and ball $=3v.\sqrt{\frac{2h}{g}}$
Comparing this with given distance we get
$n=3$