VITEEE 2013 Paper

Let two balls collide at a height s from the ground after t second when second ball is thrown upwards.
∴ Time taken by first ball to reach the point of collision = (t + 2) s
s = 39.2 (t + 2) +

1

2

(- 9.8) (t+2)2
= 39.2 (t + 2)-4.9 (t+2)2 ...(i)
For second ball
s = 39.2t +

1

2

(−9.8)t2
= 39.2t - 4.9 t2 ... (ii)
From eqs. (i) and (ii)
39.2 (t + 2) - 4.9 (t+2)2 = (39.2) t - 4.9 t2 On solving we get, t = 3s
FromEq. (ii),
s = 39.2x 3-4.9 x (3)2 = 117.6-44.1 =73.5 m