Question

If a projectile is shot at an angle of `pi/6` and at a velocity of `32 m/s`, when will it reach its maximum height??

Answer 1

`h_m=13,048 m " maximum height"`

Explanation:

`theta =cancel(pi)/6* 180/cancel(pi)=30^o`
`t=w_i*sin theta/g`
`t=32*sin 30`
`t=32*0,5=16 sec" time to maximum height"`
`h_m=(v_i^2*sin^2 30)/(2*g)`
`h_m=(32^2* 0,5^2)/(2*9,81)=(1024*0,25)/(19,62)`
`h_m=256/(19,62)`
`h_m=13,048 m " maximum height"`