Question

a rock with a mass of 2.3 kg is dropped from a height of ten meters. the rock falls for 1.43 seconds before it hits the ground.
a. what is the acceleration of the rock as it falls?
b. what force causes this acceleration?
c. determine the rock’s velocity as it hits the ground.
d. what is the rock’s momentum as it hits ground?

Explanation:

Part a

Step1: Recall acceleration formula

Acceleration \( a=\frac{\Delta v}{\Delta t} \). For free fall, initial velocity \( u = 0\,\text{m/s}\). We can also use the kinematic equation \( h=ut+\frac{1}{2}at^{2} \), but since \( u = 0\), \( h=\frac{1}{2}at^{2}\), so \( a=\frac{2h}{t^{2}} \).

Step2: Substitute values

Given \( h = 10\,\text{m}\), \( t=1.43\,\text{s}\). Plug into \( a=\frac{2h}{t^{2}}=\frac{2\times10}{(1.43)^{2}}=\frac{20}{2.0449}\approx9.78\,\text{m/s}^2\) (close to \( g = 9.8\,\text{m/s}^2\), gravitational acceleration).

The acceleration of a falling object near Earth's surface is due to the force of gravity (gravitational force) pulling the object towards the Earth.

Step1: Recall velocity formula

For free fall, \( v=u + at \). Initial velocity \( u = 0\,\text{m/s}\), \( a\approx9.8\,\text{m/s}^2\), \( t = 1.43\,\text{s}\).

Step2: Calculate velocity

\( v=0+9.8\times1.43 = 14.014\,\text{m/s}\) (or using \( v=\sqrt{2gh}=\sqrt{2\times9.8\times10}=\sqrt{196}=14\,\text{m/s}\), which matches).

Answer:

Approximately \( 9.8\,\text{m/s}^2 \) (or \( 9.78\,\text{m/s}^2 \) from calculation)

Part b