Question

select the correct answer.
a rock is dropped from a height of 2.5 meters. what is its velocity when it reaches the ground?
a. 3.5 meters/second
b. 7.0 meters/second
c. 12 meters/second
d. 14 meters/second
e. 49 meters/second

Explanation:

Step1: Identify the physics formula

We use the kinematic equation for free - fall: \(v^{2}=u^{2}+2gh\), where \(u = 0\) (initial velocity, since the rock is dropped), \(g=9.8\ m/s^{2}\) (acceleration due to gravity), and \(h = 2.5\ m\) (height).

Step2: Substitute the values into the formula

Substitute \(u = 0\), \(g = 9.8\ m/s^{2}\) and \(h=2.5\ m\) into \(v^{2}=u^{2}+2gh\). We get \(v^{2}=0 + 2\times9.8\times2.5\).

First, calculate \(2\times9.8\times2.5=49\), so \(v^{2}=49\). Then take the square root of both sides: \(v=\sqrt{49} = 7.0\ m/s\) (we take the positive root because we are interested in the magnitude of the velocity when it reaches the ground, and the direction is downward).

Answer:

B. 7.0 meters/second