Two balls of equal masses are thrown along the same vertical direction at an interval of ` 2 seconds`, with the same initial velocity of 1 45 ms^(-1)`. Then these balls collide at a height of (use g=10 ms^(-2)`

Options

A. (a) ` 54 . 2 m
B. (b) ` 96. 25 m
C. (c ) ` 16 . 7 m`
D. (d) ` 217 .0 m`

Answer

B

Explanation

Let the two balls collide at a deight (S) from the ground after (t) second when when second ball is thrown upwards. Therefore, the time taken by is first ball to reach the motion the first ball. We have
` S= 45 (t +2) + 1/2 (-10) (t+2)^2`
`=45 t +1/2 (-10) t^2=45 t- 5 t^2` ....(i)`
From (i) and (ii),
` 45 (t+2) - 5 (t-2)^2 = 45 t- 5 t^@` ltbRgt On soving we get ` t= 7//2 s``
From (ii), ` S= 45 xx 7//2 -5 xx (7//2)^2`
From (ii), ` S = 45 xx 7//s - 5 xx (7//2)^2`
` = 157.5 - 61.25 = 96. 25 m`.