A ball is projected from O with an initial velocity 700 cm/s in a direction 370 above the horizontal. A ball B, 500 cm away from O on the line of the initial velocity of A, is released from rest at the instant A is projected. The height through which B falls, before it is hit by A and the direction and magnitude of the velocity A at the time of impact will respectively be-[given g=10m/s2,sin 370 =0.6 and cos 370 =0.8]

The situation is shown in fig                             (a)      Let the ball collide after t sec from fig.OC=OBcos⁡37∘=500cos⁡37∘=500×0.8=400cm….(1) Horizontal velocity =700×cos⁡37∘∴OC=700×cos⁡370×t=700×0.8×t=560t … From eqs. (1) and (2) 560t=400 or t=(5/7)sec . h=(1/2)gt2=(1/2)×1000×(5/7)2/255.1cm (b) Let at the time of impact, vx  and  vy be the horizontal and vertical velocities respectively, then vx=700×cos⁡37∘=700×0.8=560cm/s and vy=−700×sin⁡37∘+1000×(5/7)=−700×0.6+(5000/7)=−420+714.3               =+294.3cm/secdownward Velocity of the ball at the time of collision v=vx2+vy2 ∴v=[(560)2+(294.3)2]=632.6 cm/sec Again tanθ=vyvx=294.3560 or θ=tan–1 294.3560=27043'