Question

a soccer ball is kicked horizontally off a 13.4 m high hill at 13.3 m/s. how far does it travel horizontally before it hits the ground?

Explanation:

Step1: Find the time of flight

The vertical motion of the ball is a free - fall motion. The vertical displacement \(y = 13.4\ m\), the initial vertical velocity \(u_y=0\ m/s\), and the acceleration \(a = g=9.8\ m/s^{2}\). Using the equation \(y = u_y t+\frac{1}{2}at^{2}\), substituting \(u_y = 0\), we get \(y=\frac{1}{2}gt^{2}\). Then \(t=\sqrt{\frac{2y}{g}}\).

$$t=\sqrt{\frac{2\times13.4}{9.8}}$$

$$t=\sqrt{\frac{26.8}{9.8}}\approx1.65\ s$$

Step2: Find the horizontal distance

The horizontal motion is a uniform - motion with constant velocity \(v_x = 13.3\ m/s\). Using the formula \(x = v_x t\).
Substitute \(v_x=13.3\ m/s\) and \(t = 1.65\ s\) into the formula:

$$x=13.3\times1.65 = 21.945\approx21.9\ m$$

Answer:

The soccer ball travels approximately \(21.9\ m\) horizontally before it hits the ground.