Question

A projectile is shot at a velocity of `15 m/s` and an angle of `pi/6`. What is the projectile's peak height?

Answer 1

`h_m~=2,867 m`

Explanation:

`alpha =pi/6*180/pi =30^o`
`sin 30^o=0,5`
`h_m=(v_i^2*sin^2 alpha)/(2*9,81)" maximum height"`
`h_m=(225*0,25)/(19,62)`
`h_m~=2,867 m`