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

Answer-

At 15.91° the ball was thrown.

Solution-

In the question it is given that the ball was released and caught from the same height.

We know that,

$R=\frac{u^{2}\sin 2\Theta}{g}$

Where,

R = Range of the projectile = 7.75 m,

u = Initial speed = 12 m/s

θ = Angle at which it was released

g = Acceleration due to gravity = 9.8 m/s²

Putting all the values,

$\Rightarrow 7.75 = \frac{12^{2}\sin 2\Theta}{9.8}$

$\Rightarrow \sin 2\Theta=\frac{7.75 \times 9.8}{144} =0.5274$

$\Rightarrow 2\theta=\sin^{-1} 0.5274=31.83 \ or \ 148.17$

$\Rightarrow \theta=\frac{31.83}{2} \ or \ \frac{148.17}{2} =15.91 \ or \ 74.08$

$\Rightarrow \theta=15.91^{\circ} \ (smaller \ angle)$