Question

If a projectile is shot at an angle of `(pi)/8` and at a velocity of `3 m/s`, when will it reach its maximum height?

Answer 1

Time `t=(3*sin(pi/8))/9.8=0.1171479895" "`second

Explanation:

For projectiles fired at angle `theta` with initial velocity `v_0`

`y=v_0 sin theta* t+1/2*g*t^2`

We are after the maximum height `y_max`.
We differentiate `y` with respect to `t` then equate `dy/dt=0` because we maximizing height `y`

`y=v_0 sin theta* t+1/2*g*t^2`

`dy/dt=v_0 sin theta *dt/dt+1/2*g*2*t*dt/dt`

`dy/dt=v_0 sin theta *1+1/2*g*2*t*1`

`dy/dt=v_0 sin theta+g*t`

`dy/dt=0`

`v_0 sin theta+g*t=0`

Solving for t

`t=(-v_0 sin theta)/g=(-3*sin 22.5^@)/(-9.8)=0.1171479895" "`second

God bless...I hope the explanation is useful.