Question

If a projectile is shot at an angle of `(2pi)/3` and at a velocity of `5 m/s`, when will it reach its maximum height?

Answer 1

time taken to reach maximum height = `0.4 s` (approx)

Explanation:

The velocity vector of the projectile has a x- component and y - component .
since we are dealing with height , we just need the y-component .
velocity in `y` direction = `5 sin ((2pi)/3) ms^-1`
=`(5sqrt 3) /2 ms^-1`

now using initial velocity `u = (5 sqrt3 )/2 ms^-1` ,

final velocity `v = 0 ms^-1`, acceleration due to gravity `a = -9.8 ms^-2` .

we have `v = u + at`

`=> 0 = (5sqrt3) /2 + -9.8 t`

`=> 9.8 t = (5sqrt3) /2`

`=> t = (5sqrt3) /(2×9.8)`

`=> t= 0.4 s` (approx)