A body is projected at an angle other than `90^(@)` with the horizontal with same velocity. If the time of ascent of the body is ls, then the masimum height it can reach is (Take, `g= 10 ms^(-2)`)

Options

A. 5 m
B. 10 m
C. `2.5` m
D. 75 m

Answer

A

Explanation

Given, angle of projection , `theta = 90^(@)` and time of ascent, `t = ls.`
Since, time of ascent of a projectile body is half of
the time of height `T = 2xx 1 = 2 s `
Equation of motion,
`s = ut + 1/2 gt ^(2)`
where, s = 0 for complete flight.
and gravitational acceleration, `g = 10 m//s^(2)`
putting the given values in above equation, we get
`0 = u xx 2 - 10/2 xx 2^(2)`
` u = 10 m //s`
Now, the maximum height,
`h= ut - 10/2 t^(2)` where, `t = ls` (time of ascent)
`h = 10 xx 1 - 5 xx 1 `
`rArr h = 5 m`
Hence, the correct option is (a).