Question

questions

1. a ball is dropped from a cliff and hits the ground 3.2 seconds after. how high up was the cliff?

Explanation:

Step1: Identify the kinematic equation

The equation for vertical displacement \(y - y_0=v_0t+\frac{1}{2}at^{2}\). Since the ball is dropped, \(v_0 = 0\) (initial velocity), \(a=-g=- 9.81\ m/s^{2}\) (acceleration due to gravity), and we want to find the magnitude of \(y - y_0\) (height of the cliff \(h\)). The equation simplifies to \(h=\frac{1}{2}gt^{2}\) (taking magnitude as height is positive).

Step2: Substitute the values

Substitute \(g = 9.81\ m/s^{2}\) and \(t = 3.2\ s\) into the formula \(h=\frac{1}{2}\times9.81\times(3.2)^{2}\).
First calculate \((3.2)^{2}=10.24\). Then \(\frac{1}{2}\times9.81\times10.24 = 4.905\times10.24\).
\(4.905\times10.24=(5 - 0.095)\times10.24=5\times10.24-0.095\times10.24\)
\(5\times10.24 = 51.2\), \(0.095\times10.24=(0.1 - 0.005)\times10.24=1.024-0.0512 = 0.9728\)
\(51.2-0.9728 = 50.2272\approx50.2\ m\)

Answer:

The height of the cliff is approximately \(50.2\ m\)