Question

A projectile is shot from the ground at an angle of `pi/6` 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

`t=(15*0,5)/(9,81)=0,76" "s`
`x~=9,87 m`
`y=h_m~=2,87 m`
`v_y=0" at the point of D"`
`v_x=12,99 m/s`

Explanation:

`alpha:pi/6=30^o" "v_i=15 m/s`
`t=v_i*sin alpha/g" 'elapsed time to the maximum height'"`
`t=(15*0,5)/(9,81)=0,76" "s`
`y=h_m=v_i*t*sin alpha-1/2*g*t^2`
`y=h_m=15*0,76*0,5-1/2*9,81*(0,76)^2`
`y=h_m=5,7-2,83`
`y=h_m~=2,87 m`
`x=v_i*t*cos alpha`
`x=15*0,76*0,866`
`x~=9,87 m`
`v_y=0" at the point of D"`
`v_x=v_i*cos alpha`
`v_x=15*0,866`
`v_x=12,99 m/s" The x component of velocity is not change"`