Question

two balls are at the same height and released at the same time. one ball is dropped and hits the ground 5 s later. the other initially moves horizontally. when does the second ball hit the ground? how far does it travel horizontally?

Explanation:

Step1: Analyze vertical motion

The vertical - motion of both balls is free - fall. The initial vertical velocity of both balls is \(v_{0y}=0\ m/s\), and they experience the same acceleration \(a = g=9.8\ m/s^{2}\) (downward). Since they start from the same height \(h\), using the equation \(h = v_{0y}t+\frac{1}{2}gt^{2}\), and \(v_{0y} = 0\), the time of fall \(t=\sqrt{\frac{2h}{g}}\) is the same for both. So the second ball also hits the ground in \(t = 5\ s\).

Step2: Analyze horizontal motion

The second ball has a constant horizontal velocity \(v_x=13\ m/s\) (because there is no horizontal acceleration in the absence of air - resistance). Using the formula \(x = v_x t\), where \(v_x = 13\ m/s\) and \(t = 5\ s\).

$$x=13\times5$$

Answer:

The second ball hits the ground in \(5\ s\) and travels horizontally a distance of \(65\ m\).