Question

a projectile is projected with speed u at an angle theta from horizontal. it collides with the wall elastically and then fall on the ground at a point vertically below the highest point as shown. the value of Y is​

Answer 1

This is very interesting question.

a projectile is projected with speed u at angle $\theta$ from the horizontal, it collides with the wall elastically and then fall on the ground at a point vertically below the highest point. here only the path of projectile is changing , the type of motion is exactly same. I mean, time of ascending =total time taken to moves after reaching the ground.

if we assume time of ascending is 2t.

then, time taken to reach the ground after that is also 2t.

but here also time taken by projectile from maximum height to wall = time taken by projectile from wall to ground [ because collision is elastic so, there doesn't effect on path and motion]

so, A final diagram is shown in figure.

using formula, S = s' + uT + 1/2 aT²

S = -H , u = 0, s' = 0 [initial position of particle with respect to frame of reference [ vertically component at maximum height is zero ]

- H = -1/2 g(2t)² = 2gt² ......(1)

now again, S = -y, s' = -H , a = g , T = t

-y = -H + 1/2 gt²

or, y = H - 1/2 gt²

from equation (1),

y = H - H/4 = 3H/4