Question

A projectile is shot from the ground at an angle of `(5 pi)/12` and a speed of `15 m/s`. Factoring in both horizontal and vertical movement, what will the projectile's distance from the starting point be when it reaches its maximum height?

Answer 1

`"1) "x=v_i*t*cos alpha " horizontal movement"`
`"2) "y=v_i*t*sin alpha-1/2*g*t^2" vertical movement"`
`"3) "x~=5,734m`

Explanation:

`v_i:15 m/s " initial velocity"`
`alpha:(5cancel(pi))/12*180/cancel(pi)=75^o" sin75~=0,966 cos75~=0,259"`
`v_x=v_i*cos75 " x component of initial velocity"`
`v_y=v_i.sin75 " y component of initial velocity"`
`t_m=v_i*sin alpha/g "elapsed time to reach to max height"`
`h_m=(v_i^2*sin^2 alpha)/2*g" maximum height"`
`"1) "x=v_i*t*cos alpha " horizontal movement"`
`"2) "y=v_i*t*sin alpha-1/2*g*t^2" vertical movement"`
`"3) "t_m=(15*sin75)/(9,81~)=1,477 s`
`x=15*cos75*1,477`
`x~=5,734m`