Question

A projectile is shot from the ground at an angle of `(5 pi)/12` and a speed of `5/8 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

`v_x=0,16 " m/s"`
`v_y=0,60" m/s"`
`"x-component of velocity at any time:" v_x=v_i*cos alpha`
`"y-component of velocity at any time:"v_y=v_i*t*sin alpha-g*t`
`x_m=0,02" "m`
`t=0,06" s"`

Explanation:

`"x-component of the initial velocity : "v_x=v_*cos alpha`

`v_x=5/8*cos ((5pi)/12)`

`v_x=0,16 " m/s"`

`"y-component of the initial velocity : "v_y=v_i*sin alpha`

`v_y=5/8*sin((5pi)/12)`

`v_y=0,60" m/s"`

`"x-component of velocity at any time:" v_x=v_i*cos alpha`
`"y-component of velocity at any time:"v_y=v_i*t*sin alpha-g*t`

`"for maximum x: "x_m=(v_i^2*sin2 alpha)/g`

`x_m=((5/8)^2*sin 2*(5 pi)/12)/(9,81)`

`x_m=(25*sin((5pi)/6))/(64*9,81)=(25*0,5)/(64*9,81)`

`x_m=0,02" "m`

`"for time to maximum height : "t=(v_i*sin alpha)/g`

`t=(5*sin((5pi)/12))/(8*9,81)=(4,8296291314)/(78,48)`

`t=0,06" s"`