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Chapter 2: Problem 55

A rock is dropped from a 100 -m-high cliff. How long does it take to fall \((a)\) the first \(50.0 \mathrm{~m}\) and \((b)\) the second \(50.0 \mathrm{~m}\) ?

Short Answer

Expert verified
The time taken to fall the first 50m is \(t1=\sqrt{\frac{2 \times 50}{9.8}}\) seconds. The time taken to fall the second 50m is the total time \(t_{\mathrm{total}}=\sqrt{\frac{2 \times 100}{9.8}}\) seconds minus \(t1\).

Step by step solution

01

Calculate the time taken to fall the first 50m

Use the formula \(t=\sqrt{\frac{2d}{g}}\) where \(g=9.8 \mathrm{ms}^{-2}\) and \(d=50 \mathrm{m}\). The time taken to fall the first 50m (\(t1\)) can be calculated as: \(t1=\sqrt{\frac{2 \times 50}{9.8}}\)
02

Calculate the time taken to fall the total 100m

Again use the formula \(t=\sqrt{\frac{2d}{g}}\). This time, \(d=100 \mathrm{m}\). The total time taken to fall 100m (\(t_{\mathrm{total}}\)) can be calculated as: \(t_{\mathrm{total}}=\sqrt{\frac{2 \times 100}{9.8}}\)
03

Calculate the time taken to fall the second 50m

The time taken to fall the second 50m (\(t2\)) is simply the total time taken to fall 100m minus the time taken to fall the first 50m. So, \(t2=t_{\mathrm{total}} - t1\).

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A rocket ship in free space moves with constant acceleration equal to \(9.8 \mathrm{~m} / \mathrm{s}^{2} .(a)\) If it starts from rest, how long will it take to acquire a speed one-tenth that of light? \((b)\) How far will it travel in so doing? (The speed of light is \(3.0 \times 10^{8} \mathrm{~m} / \mathrm{s}\).)

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