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Chapter 2: Problem 45

A diver drops from a 9.21-m high cliff. (a) At what speed does she enter the water? and (b) how long is she in the air?

Short Answer

Expert verified
The diver enters the water at a speed of approximately 13.46 m/s and is in the air for approximately 1.37 seconds.

Step by step solution

01

Calculate the speed

The speed of an object in free fall after time \(t\) is given by \(v = gt\). But, we do not have the time here, we have the distance, which in motion physics is related by the third equation of motion for displacement \(d = 0.5gt^{2}\). We could apply this principle, rearrange for time and substitute in the first equation. However, there's a shortcut using another version of an equation from kinematics for objects dropped from rest, which is \(v = \sqrt{2gd}\)}. Given that \(d = 9.21 m\) and \(g = 9.8 m/s^{2}\), substituting these in the equation gives \(v = \sqrt{2 * 9.8 * 9.21} = 13.46 m/s\). Therefore, the diver enters the water at a speed of 13.46 m/s.
02

Calculate the time

From the distance equation \(d = 0.5gt^{2}\), we can find time by rearranging the equation and solving for \(t\) which gives us \(t = \sqrt{2d / g}\). Hence substituting for \(d = 9.21 m\) and \(g = 9.8 m/s^{2}\) in the equation gives \(t = \sqrt{2*9.21 / 9.8} = 1.37 s\). Therefore, the diver is in the air for approximately 1.37 seconds.

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