Question

a student drops a ball from a height of 80.0m. if the ball increases speed at a uniform rate of 9.81m/s², determine all unknowns and answer the following questions. how long did the ball remain in the air? what was the balls speed just before striking the ground?

Explanation:

Step1: Find the time the ball remains in the air

Use the equation \(h = v_0t+\frac{1}{2}at^2\). Since the ball is dropped, \(v_0 = 0\). The equation becomes \(h=\frac{1}{2}at^2\). We know \(h = 80.0m\) and \(a=9.81m/s^{2}\).

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Step2: Calculate the value of \(t\)

$$ t=\sqrt{\frac{160}{9.81}}\approx 4.04s $$

Step3: Find the ball's speed just before striking the ground

Use the equation \(v = v_0+at\). Since \(v_0 = 0\), \(v = at\). Substitute \(a = 9.81m/s^{2}\) and \(t\approx4.04s\)

$$ v=9.81\times4.04 = 39.6m/s $$

Answer:

The ball remains in the air for approximately \(4.04s\) and the ball's speed just before striking the ground is approximately \(39.6m/s\)