Question

If a projectile is shot at an angle of `(pi)/8` and at a velocity of `35 m/s`, when will it reach its maximum height?

Answer 1

First find the vertical velocity `v_y = vsintheta`
Where `theta=pi/8 or 45/2`
Then use the kinematic formula:
`v_f^2= v_i ^2+2ad`
Where `v_f=0; a=g=9.81m/s^2`
Note as you substitute Gravity is in the opposite direction thus negative. Solve and you have it

`d=sqrt(v_y^2/(2g))`
`=sqrt(((v^2sin^2theta)/(2g))`