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/ s2. 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 450 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