A rugby player passes the ball 7.75 m across the field, where it is caught at the same height as it left his hand. At what angle was the ball thrown if its initial speed was 12.0 m/s, assuming that the smaller of the two possible angles was used?
Accepted Solution
A:
Given that the "launch" and "impact" heights are the same, we shall use the range equation and the time of flight equation. (a)(b)range x=V²sin (2θ)/g 7.75m=(12 m/s)sin (2θ)/ 9.8m/s² sin (2θ)=0.5274 hence: 2θ=arcsin 0.5274=31.83° or 2θ=180-31.83=148.17° hence the smaller angle will be: 31.83/2=15.915~15.92° and the larger angle will be: 148.17/2=74.085°