Question

A projectile is shot at a velocity of `2 m/s` and an angle of `pi/6`. What is the projectile's peak height?

Answer 1

`h_p~=0,051 m`

Explanation:

`"The projectile's peak height can be calculated using the formula:"`
`h_p=(v_i^2*sin^2 alpha)/(2*g)`
`v_i" :initial velocity of object"`
`alpha" :angle of projectile"`
`g" : acceleration of gravity"`
`alpha=pi/6*180/pi=30^o`
`sin 30=0,5`
`sin^2 30=0,25`
`h_p=(2^2*0,25)/(2*9,81)`
`h_p=(2*0,25)/(9,81)`
`h_p~=0,051 m`