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? 41.9 m 137 m 119 m 238 m

Explanation:

Step1: Find the initial vertical velocity

At the peak of the motion, the vertical velocity \(v_y = 0\). Using the kinematic equation \(v_y^2=v_{0y}^2 - 2gh\) (where \(v_y = 0\), \(g = 9.8\ m/s^2\), and \(h = 17.2\ m\)).

$$0=v_{0y}^2-2\times9.8\times17.2$$

$$v_{0y}^2 = 2\times9.8\times17.2$$

$$v_{0y}=\sqrt{2\times9.8\times17.2}\approx18.4\ m/s$$

Since \(v_{0y}=v_0\sin\theta\) (\(\theta = 30^{\circ}\)), then \(v_0=\frac{v_{0y}}{\sin\theta}=\frac{18.4}{\sin30^{\circ}} = 36.8\ m/s\)

Step2: Find the time of flight

The time to reach the peak \(t_{up}=\frac{v_{0y}}{g}=\frac{18.4}{9.8}\approx1.88\ s\). The total time of flight \(T = 2t_{up}=2\times1.88 = 3.76\ s\)

Step3: Find the horizontal velocity and horizontal distance

The horizontal velocity \(v_{0x}=v_0\cos\theta=36.8\cos30^{\circ}\approx31.9\ m/s\). Using the formula \(x = v_{0x}T\), \(x=31.9\times3.76\approx119\ m\)

Answer:

119 m