A body is projected vertically upwards with some velocity `u`. It is at the same point at `t = 6 s` as was at `t = 4 s`. Find the value of `u`.

Options

A. `30 m//s`
B. `40 m//s`
C. `20 m//s`
D. `50 m//s`

Answer

D

Explanation

At `t_1 = 4s` and `t_2 = 6 s`, body is at height `h`
So `h = ut_1 - ((1)/(2)) g t_1^2`
and `h = ut_2 - ((1)/(2)) g t_2^2`
Solve to get `u = (1)/(2) g(t_1 + t_2)`
=`(1)/(2) xx 10(4 + 6) = 50 m//s`
Time of flight : `T = t_1 + t_2`
`(2 u)/(g) = T rArr (2 u)/(10) = 10 rArr u = 50 m//s`.