Question

A projectile is shot at a velocity of `21 m/s` and an angle of `pi/6`. What is the projectile's peak height?

Answer 1

`h_m=1,124 m`

Explanation:

`v_i=21 m/s" "alpha=pi/6" "sin (pi/6)=0,5`
`h_m=(v_i^2*sin^2 alpha)/(2g)`
`h_m=(21^2*(0,5)^2)/(2*9,81)`
`h_m=(22,05)/(19,62)`
`h_m=1,124 m`