Question

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?

Answer 1

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°

Answer 2

The smaller of the two possible angles used is;

θ = 15.92°.

We are given;

Range of pass; R = 7.75 m

Speed; u = 12 m/s

Formula for range is;

R = [u²sin (2θ)]/g

Where; θ is the angle at which the ball was thrown.

Thus;

7.75 = 12² × (sin(2θ))/9.8

(sin(2θ)) = (7.75 × 9.8)/12²

(sin(2θ)) = 0.5274

2θ = sin^(0.5274)

2θ = 31.83

θ = 31.83/2

θ = 15.92°

Now, if we assume that this angle is to the vertical, it means the angle to the horizontal is;

90 - 15.92 = 74.08°

In conclusion, the two possible angles are 15.92° and 74.08°.

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