A ball is thrown up, it reaches a maximum height anf then comes down. IF ` t_1` and `t_2` (t_2 gt t_1)` ar the time that the ball takes to be at a particular height then the time taken by the ball to reach the highest point is .

Options

A. (a) ` (t_1+t_2)`
B. (b) ` (t_2 -t_1)`
C. (c ) ` (t_2 -t_1)//2`
D. (d) ` (t_2 +t_1)//2`

Answer

D

Explanation

Let (S) be the height of a particular point where the ball croosses in time ` t_1 ` and t_2` seconds while going upwards and coming downwards. If (u) is the initial velocity of projection of ball, then
` S= ut-1 - 1/2 gt_1^2 = ut^2 - 1/2 gt_2^@`
or ` u ( t_1 -t_2) = 1/2 g (t_2^2 -t_1^2)`
or ` u = 1/2 g (t- +t-10 . if (T) si the time taken by ball to reach to its highest then using the relation
`v = u+ at, we have ` 0 = u + ( - g) T` ltbRgt or ` T= u /g = 1/2 g ( t_2- t_1) //g = 1/2 (t_2-t_1)`.