Question

If a projectile is shot at an angle of `pi/4` and at a velocity of `16 m/s`, when will it reach its maximum height??

Answer 1

1.15s

Explanation:

A projectile reaches its maximum height when vertical velocity is zero.

Initial velocity, u = `16cos(pi/4)=16*sqrt2/2=8sqrt2`

Taking acceleration as `a=-9.8ms^-1` then:

`v = u + at`
`0=8sqrt2-9.8t`
`t=(8sqrt2)/9.8=1.15s` (3sf)