Question

1. a stunt car traveling at 20 m/s flies horizontally off a cliff and lands 39.2 m from the base of the cliff. how tall is the cliff? (round to the nearest tenths place.)

Explanation:

Step1: Calculate the time of flight

The horizontal motion is uniform. Using the formula \(x = v_{0x}t\), where \(x = 39.2\space m\) and \(v_{0x}=20\space m/s\).

$$t=\frac{x}{v_{0x}}=\frac{39.2}{20}=1.96\space s$$

Step2: Calculate the height of the cliff

The vertical motion is free - fall (\(v_{0y} = 0\space m/s\)). Using the formula \(y=v_{0y}t+\frac{1}{2}gt^{2}\), where \(g = 9.8\space m/s^{2}\) and \(t = 1.96\space s\)

$$y=0\times t+\frac{1}{2}\times9.8\times(1.96)^{2}$$

$$y = 4.9\times3.8416$$

$$y=18.82384\approx18.8\space m$$

Answer:

\(18.8\space m\)