A large , heavy box is sliding without friction down a smooth plane of inclination `theta` . From a point `P` on the bottom of the box , a particle is projected inside the box . The initial speed of the particle with respect to the box is `u` , and the direction of projection makes an angle `alpha` with the bottom as shown in Figure .
Find the distance along the bottom of the box between the point of projection `p` and the point `Q` where the particle lands . ( Assume that the particle does not hit any other surface of the box . Neglect air resistance .

Answer

`(3.75hati+6.25 ghatj)m//s`

Explanation

Let after time ‘t’ the stone hits the object and be the angle which the velocity vector makes with horizontal
From question, we have following information Vertical displacement of stone is 1.25 m
Using `s_(y)=u_(y)t+1/2a_(y)t^(2)`
Therefore 1.25 `=(u sin theta)t-1/2 gt^(2) or (u sin theta)t=1.25 +5t^(2) ...(i)`
Horizontal displacement of stone = 3 + displacement of object A.
Therefore `(u cos theta)t=3+1/2 at^(2)`
We have `a=1.5 m//s^(2), "Hence "(u cos theta)t=3+0.75 t^(2) .....(ii)`
Horizontal component of velocity (of stone) = vertical component (because velocity vector is inclined at `45^(@)` with horizontal).
Therefore `(u cos theta)=gt-(u sin theta) ........(iii)`
The right-hand side is written because the stone is in its downward motion. Therefore `gt gt u sin theta`
Multiplying equation (iii) with t can write. `(u cos theta)t+ (u sin theta)t=10t^(2) ......(iv)`
Now doing (iv)-[(iii)+(i)]
It gives `4.25t^(2)-4.25 or t=1s " Substituting t=1 s in (i) and (ii) we get "`
`u sin theta =6.25 m //s or v_(y)=6.25 m and u cos theta=3.75 m//s or u_(x)=3.75 m//s`
Now, we can write `u=u_(x)hati+u_(y)hati or u=(3.75 hati+6.25 hatj)m//s`