Question

5 sean is a kicker for a college football team and is practicing kicking for an upcoming game. on one attempt, he kicks the football with a velocity of 21.0 meters per second in the positive direction, at an angle of 32.7° above horizontal. assuming the ball lands back on the field, what is the range of seans kick? 34.4 m 40.9 m 40.0 m 38.9 m

Explanation:

Step1: Find the horizontal and vertical components of velocity

The initial velocity \(v = 21.0\space m/s\) and the angle \(\theta=32.7^{\circ}\).
The horizontal component \(v_{x}=v\cos\theta\) and the vertical component \(v_{y}=v\sin\theta\).
\(v_{x}=21\cos(32.7^{\circ})\approx21\times0.843 = 17.7\space m/s\)
\(v_{y}=21\sin(32.7^{\circ})\approx21\times0.539 = 11.32\space m/s\)

Step2: Find the time of flight

Using the vertical motion equation \(y = v_{y}t-\frac{1}{2}gt^{2}\). Since the ball lands back on the field (\(y = 0\)), \(0=v_{y}t-\frac{1}{2}gt^{2}\). Factoring out \(t\) gives \(t( v_{y}-\frac{1}{2}gt)=0\). One solution is \(t = 0\) (initial time), and the other is \(t=\frac{2v_{y}}{g}\) (where \(g = 9.8\space m/s^{2}\)).
\(t=\frac{2\times11.32}{9.8}\approx2.31\space s\)

Step3: Find the range

The range \(R\) is given by \(R = v_{x}t\).
Substitute \(v_{x}=17.7\space m/s\) and \(t = 2.31\space s\)
\(R=17.7\times2.31\approx40.9\space m\)

Answer:

40.9 m