Question

A projectile is shot from the ground at an angle of `( pi)/3` and a speed of `12 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

Resolve velocity into x and y components:
`v_x=12cos(pi/3)=6.00ms^-1`
`v_y=12sin(pi/3)=10.4ms^-1`

Maximum height `=> v_y=0`

Constant acceleration, given `v` and `a` so use this kinematic equation:
`v_(y)^2-v_(y0)^2=2a_y(s_(y)-s_(y0))`
Solve for `s_y`:
`0-10.4^2=2*-9.81*s_y => s_y=5.50m`

Solve for `t`:
`v_y-v_(y0)=a_y(t-t_0)=>t=1.06s`
Air resistance assumed negligible `=>a_x=0`
Solve for `s_x`:
`s_x-s_(x0)=v_(x0)(t-t_0)+1/2a_x(t-t_0)^2`
`s_x=6*1.06=6.36m`

Total distance travelled = `sqrt(s_x^2+s_y^2)=8.41m`