Question

A projectile is shot at a velocity of `3 m/s` and an angle of `pi/8`. What is the projectile's peak height?

Answer 1

`h_(peak)=0,00888` `"meters"`

Explanation:

`"the formula needed to solve this problem is:"`
`h_(peak)=(v_i^2*sin^2 theta/(2*g))`
`v_i=3 m/s`
`theta =180/cancel(pi)*cancel(pi)/8`
`theta =180/8`
`sin theta=0,13917310096`
`sin^2 theta=0,0193691520308`
`h_(peak)=3^2*(0,0193691520308)/(2*9,81)`
`h_(peak)=9*(0,0193691520308)/(19,62)`
`h_(peak)=0,00888` `"meters"`