An object `A` is kept fixed at the point ` x= 3 m` and `y = 1.25 m` on a plank `p` raised above the ground . At time ` t = 0 ` the plank starts moving along the `+x` direction with an acceleration ` 1.5 m//s^(2) `. At the same instant a stone is projected from the origin with a velocity `vec(u)` as shown . A stationary person on the ground observes the stone hitting the object during its downward motion at an angle ` 45(@)` to the horizontal . All the motions are in the ` X -Y `plane . Find ` vec(u)` and the time after which the stone hits the object . Take ` g = 10 m//s`

Answer

1s

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`