Question

If a projectile is shot at an angle of `(pi)/8` and at a velocity of `21 m/s`, when will it reach its maximum height?

Answer 1

The time is `=0.82s`

Explanation:

Resolving in the vertical direction `uarr^+`

The initial velocity is `u_0=(21)sin(1/8pi)ms^-1`

At the greatest height, `v=0ms^-1`

The acceleration due to gravity is `a=-g=-9.8ms^-2`

The time to reach the greatest height is `=ts`

Applying the equation of motion

`v=u+at=u- g t`

The time is `t=(v-u)/(-g)=(0-(21)sin(1/8pi))/(-9.8)=0.82s`