Question

a rugby player passes the ball 6.75 m across the field, where it is caught at the same height as it left his hand. at what angle θ was the ball thrown if its initial speed was 12.0 m/s, assuming that the smaller of the two possible angles was used? what other angle θ₂ gives the same range? for the pass using the smaller of the two angles, how long t₁ is the ball in the air? question credit: openstax college physics

Explanation:

Step1: Recall the range formula for projectile motion

The range \( R \) of a projectile launched with initial speed \( v_0 \) at angle \( \theta \) is given by \( R=\frac{v_0^2\sin(2\theta)}{g} \), where \( g = 9.8\space m/s^2 \) is the acceleration due to gravity. We know \( R = 6.75\space m \), \( v_0=12.0\space m/s \). First, solve for \( \sin(2\theta) \).

Rearranging the formula: \( \sin(2\theta)=\frac{Rg}{v_0^2} \)

Substitute the values: \( \sin(2\theta)=\frac{6.75\times9.8}{12.0^2}=\frac{66.15}{144}\approx0.4594 \)

Step2: Find the two possible angles for \( 2\theta \)

We know that if \( \sin\alpha=\sin\beta \), then \( \beta = 180^\circ-\alpha \) (in degrees) or \( \beta=\pi - \alpha \) (in radians). So first, find \( \alpha = \arcsin(0.4594) \)

\( \alpha=\arcsin(0.4594)\approx27.3^\circ \) (in degrees). Then the other angle for \( 2\theta \) is \( 180^\circ - 27.3^\circ=152.7^\circ \)

Step3: Find \( \theta_1 \) and \( \theta_2 \)

For \( \theta_1 \) (smaller angle), \( 2\theta_1 = 27.3^\circ \), so \( \theta_1=\frac{27.3^\circ}{2}\approx13.6^\circ \)

For \( \theta_2 \), \( 2\theta_2 = 152.7^\circ \), so \( \theta_2=\frac{152.7^\circ}{2}\approx76.4^\circ \)

Step4: Find the time of flight \( t_1 \) for \( \theta_1 \)

The time of flight \( t \) for a projectile (when launched and landed at same height) is given by \( t=\frac{2v_0\sin\theta}{g} \)

For \( \theta_1 = 13.6^\circ \), \( \sin(13.6^\circ)\approx\sin(13.6^\circ)\approx0.234 \)

\( t_1=\frac{2\times12.0\times0.234}{9.8}=\frac{5.616}{9.8}\approx0.573\space s \)

Answer:

s:
\( \theta_1\approx\boxed{13.6^\circ} \)

\( \theta_2\approx\boxed{76.4^\circ} \)

\( t_1\approx\boxed{0.573\space s} \)