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

`u = 7.29 m//s, t = 1sec`

Explanation

Let stone hit the plank at time `t` after projection. Taking motion along y-axis
`u = u_(y)t +(1)/(2) a_(y)t^(2)`

`1.25 = u sin theta t = (1)/(2) g t^(2)`
`u sin theta t = (5)/(4) +5t^(2) ….(1)`
Taking motion along x-axis
`x = u cos theta t = 3+(1)/(2) at^(2)`
`u cos theta t = 3 +(1.5)/(2) t^(2) rArr u cos theta t = 3+(3)/(4) t^(2) ....(2)`
at striking point, `v_(y) = u_(y) +a_(y) t`
`-u cos theta = u sin theta - g t rArr u sin theta + u cos theta = g t ...(3)`
From (1) and (2)
`(ucos theta + usin theta) t = (5)/(4) +3 +5t^(2) +(3)/(4) t^(2) rArr`
`(ucos theta + usin theta ) t = (5+12 +20t^(2) +3t^(2))/(4)`
`(ucos theta +u sin theta)t = (17 +23t^(2))/(4) ...(4)`
From (3) and (4) `"gt"^(2) = (17 +23 t^(2))/(4)`
`40t^(2) = 17 +23 t^(2) rArr 17t^(2) = 17 rArr`
`t = 1sec`.
By putting the value of `t` in equation (1) and (2)
`{:(usin theta = (5)/(4)+5,,rArr,usin theta = (25)/(4),rArr),(ucostheta = 3+(3)/(4),,rArr,ucos theta = (15)/(4),):}`
`u = sqrt(((25)/(4))^(2)+((15)/(4))^(2))rArru = (1)/(2) sqrt(625+225)`
`rArr u = (sqrt(850))/(4) rArr y = 7.29 sec`.