Question

1. a girl throws a rock horizontally from the top of a cliff 98 m high, with a horizontal velocity of 27 m/s.

Explanation:

Step1: Find the time it takes for the rock to fall vertically

The vertical - motion of the rock is a free - fall motion. The height of the cliff \(h = 98\ m\), and the initial vertical velocity \(u_y=0\ m/s\). The equation for vertical displacement \(h = u_y t+\frac{1}{2}gt^{2}\), since \(u_y = 0\ m/s\), the equation simplifies to \(h=\frac{1}{2}gt^{2}\), where \(g = 9.8\ m/s^{2}\). Solving for \(t\), we get \(t=\sqrt{\frac{2h}{g}}\).
\[t=\sqrt{\frac{2\times98}{9.8}}=\sqrt{20}\approx 4.47\ s\]

Step2: Find the horizontal distance

The horizontal motion of the rock is a uniform - motion with a constant horizontal velocity \(u_x = 27\ m/s\). The horizontal distance \(x\) is given by the formula \(x = u_x t\).
\[x=27\times4.47 = 120.69\ m\]

Answer:

The rock lands approximately \(120.7\ m\) from the base of the cliff.