Question

a kicker punts a football at an angle of 30° to the field. the football reaches 17.2 m in the air at its peak. how far does the ball travel horizontally?

Explanation:

Step1: Find the initial vertical velocity

Use the formula \(v_{y}^2 = v_{0y}^2- 2gh\). At the peak, \(v_{y} = 0\). So \(v_{0y}=\sqrt{2gh}\). Given \(h = 17.2m\) and \(g=9.8m/s^{2}\), then \(v_{0y}=\sqrt{2\times9.8\times17.2}\approx18.4m/s\).

Step2: Find the initial velocity

Since \(v_{0y}=v_{0}\sin\theta\) and \(\theta = 30^{\circ}\), then \(v_{0}=\frac{v_{0y}}{\sin\theta}=\frac{18.4}{\sin30^{\circ}} = 36.8m/s\).

Step3: Find the time of flight

The time to reach the peak is \(t_{1}=\frac{v_{0y}}{g}=\frac{18.4}{9.8}\approx1.88s\). The total time of flight \(T = 2t_{1}=3.76s\).

Step4: Find the horizontal velocity

\(v_{0x}=v_{0}\cos\theta=36.8\cos30^{\circ}\approx31.9m/s\).

Step5: Find the horizontal distance

Use the formula \(x = v_{0x}T\). Substitute \(v_{0x}\approx31.9m/s\) and \(T = 3.76s\), then \(x=31.9\times3.76\approx120m\).

Answer:

The ball travels approximately \(120m\) horizontally.