## 2000 (10 Marks)

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 u as shown. A stationary person on the ground observes the stone hitting the object during its downward motion at an angle of 45° to the horizontal. All the motions are in x-y plane. Find u and the time after which the stone hits the object. (Take g = 10 m/s ^{2})

Solution : Let's work in a frame in which A is at rest, the acceleration of the ball in this frame is \( \leftarrow a = 1.5 ms^{-2} \) and \( \downarrow =g \). Let the projectile makes an angle \(\theta\) with x axis then

\(3 = u cos\theta t - \frac{1}{2}at^{2}\) ...........(1)

and \(1.25 = u sin \theta t - \frac{1}{2}gt^{2}\) ...........(2)

Also when the projectile hits the target A, this is given that the projectile makes an angle 45^{0} with the horizontal and moving in downward direction, hence as per projectile motion seen from the ground, we can write, \(u cos \theta = gt - u sin \theta\).

This gives us \(u cos \theta + u sin \theta = gt\) ..........(3)

Adding equations (1) and (2) and putting \(u cos \theta + u sin \theta = gt\) gives us t = 1s. Now put this value of t in the equations (1) and (2) and get \(u cos \theta\) = 3.75 and \(u sin \theta\) = 6.25. Get the rest, whatever you want