Question

a stone is thrown horizontally at 6.0 m/s from a cliff 13 m high. how long does it take for the stone to hit the ground (time of flight?)

Explanation:

Step1: Analyze vertical motion

The vertical motion of the stone is a free - fall motion. The initial vertical velocity \(v_{0y}=0\ m/s\), the vertical displacement \(y = 13\ m\), and the acceleration \(a = g=9.8\ m/s^{2}\). The equation for vertical displacement in free - fall is \(y=v_{0y}t+\frac{1}{2}at^{2}\). Since \(v_{0y} = 0\ m/s\), the equation simplifies to \(y=\frac{1}{2}gt^{2}\).

Step2: Solve for time \(t\)

From \(y=\frac{1}{2}gt^{2}\), we can solve for \(t\). Rearranging the formula gives \(t=\sqrt{\frac{2y}{g}}\). Substitute \(y = 13\ m\) and \(g = 9.8\ m/s^{2}\) into the formula: \(t=\sqrt{\frac{2\times13}{9.8}}=\sqrt{\frac{26}{9.8}}\approx\sqrt{2.65}\approx1.6\ s\)

Answer:

A. 1.6 s