Two balls of equal masses are thrown upwards along the same vecticla direction at an interavel of ` 2 seconsds`. With the same intial velcity of ` 39.2 m//s` . The these collede at a height of .

Options

A. (a) ` 44.1 m`
B. (b) ` 73.5 m`
C. (c ) ` 11 .6 m`
D. (d) 1 9 .0 m`

Answer

B

Explanation

Here `u = 39.2 m//s`
Let two balls collide at a height (S) from the groun after (t) seconds when second ball is thrown upwards.
:. `the time taken by first ball to teach the point of collision = ( t+ 2) sec`.
Taking motion of the first ball, we have
` S= 39.2 ( t + 2) + 1/2 (- 9.8 ) ( t+ 2)^@`
`= 39.2 (t + 2) - 4 .9 ( t+ 2)^2` ...(i)
Taking motion of second ball, we have
` S= 39.2 t = 1/2 (- 9.8 0 T^2 = 39.2 t- 4.9 t^@` ...(ii)
From (i) and (ii), we have ltbRgt ` 39.2 (t+2) - 4.9 (t + 2)^2 = 39. 2 t - 4 .9 t^2`
On solving we get ` t= 3 sec `
From (ii), ` S = 39.2 xx 2 - 4.9 (3)^2 = 117. 6 - 44.1`
`= 73.5 m`.