Question

1. a boy throws a rock straight up with an initial speed of 8 m/s. what is the rock’s speed when he catches it again? (assume no air friction, as usual.)

Explanation:

Step1: Analyze the motion

The rock is in free - fall motion. The acceleration due to gravity \(g=- 9.8\ m/s^{2}\) (taking upward as positive). The displacement \(y = 0\) when the boy catches the rock again. We use the kinematic equation \(v^{2}=v_{0}^{2}+2ay\).

Step2: Substitute values

Given \(v_{0}=8\ m/s\), \(a=-g=- 9.8\ m/s^{2}\), and \(y = 0\).
Substitute into the equation \(v^{2}=v_{0}^{2}+2ay\):

$$
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Since the direction is downward when caught (opposite to the initial upward direction), we take the negative value. But speed is a scalar quantity, so we take the magnitude.

Answer:

\(8\ m/s\)