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Chapter 2: Problem 1

Two athletes jump straight up. Upon leaving the ground, Adam has half the initial speed of Bob. Compared to Adam, Bob jumps a) 0.50 times as high. b) 1.41 times as high. c) twice as high. d) three times as high. e) four times as high.

Short Answer

Expert verified
Answer: e) Bob jumps four times as high as Adam.

Step by step solution

01

1. Write down given information

We have the initial speed of Adam and Bob, with Adam having half the initial speed of Bob. Let Va be the initial speed of Adam and Vb be the initial speed of Bob. Then: Va = 0.5 * Vb
02

2. Use kinematic equations

We know that the height (h) depends on the initial speed (v0) of a jump. The maximum height is achieved when the final velocity is zero (vf=0). Using the kinematic equation: vf^2 = v0^2 - 2gh, we can find the height by rearranging it for h: h = (v0^2) / (2g) Where g is the acceleration due to gravity, which is approximately 9.81 m/s^2.
03

3. Substitute initial speeds

Now we can find the height jumped by both Adam and Bob using their initial speeds. For Adam: ha = (Va^2) / (2g) For Bob: hb = (Vb^2) / (2g)
04

4. Compare their heights

We need to find the ratio of their heights (hb / ha), to determine by how much higher Bob jumps compared to Adam: hb / ha = ((Vb^2) / (2g)) / ((Va^2) / (2g)) The '2g' in both the numerator and the denominator cancels out, and we can substitute Va = 0.5 * Vb: hb / ha = (Vb^2) / ((0.5 * Vb)^2)
05

5. Simplify the expression

When simplifying the expression, we get: hb / ha = (Vb^2) / (0.25 * Vb^2) = 1/0.25 = 4 So, the answer is: e) Bob jumps four times as high as Adam.

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