The trapdoor swings open and the ball falls to the ground. The ball does not bounce when it hits the ground. Show that the final speed of the ball at the instant before it hits the ground is about 5 m/s.

5 m/s

Explanation:

As the ball falls to the ground, it is acted upon by a force which is due to the gravitational force. Hence, the ball falls with an acceleration called the acceleration due to gravity.

Let us assume that the acceleration due to gravity (g) = 10 m/s², Since initially, the body was at rest, hence the initial velocity (u) = 0 m/s. The height to the ground (h) = 1.3 m. Hence using the equation of motion, the final velocity (v) is:

v² = u² + 2gh

Substituting:

v² = 0² + 2(10)(1.3)

v² = 26

v = √26 = 5

v = 5 m/s

answer; (arch); good luck

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