Question

a ball is thrown horizontally from a cliff at a speed of 8 m/s. if it takes 4 seconds to hit the ground, how high is the cliff? (use acceleration due to gravity, a = g = 9.8m/s²)
a 40 m
b 60 m
c 78 m
d 100 m

Explanation:

Step1: Identify the vertical motion formula

For vertical motion under gravity (initial vertical velocity \(u_y = 0\)), the formula \(h=u_y t+\frac{1}{2}gt^{2}\) is used. Since \(u_y = 0\), the formula simplifies to \(h = \frac{1}{2}gt^{2}\).

Step2: Substitute the values

Given \(g = 9.8m/s^{2}\) and \(t = 4s\). Substitute into \(h=\frac{1}{2}\times9.8\times4^{2}\).
First calculate \(4^{2}=16\). Then \(\frac{1}{2}\times9.8\times16=(9.8\times8)\).
\(9.8\times8=(10 - 0.2)\times8=80-1.6 = 78.4\approx78m\) (approximate due to option - matching in a multiple - choice context).

Answer:

C. 78 m