Question

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

Short answer

a) The same amount of time b) 1.5 times as long c) Twice as long d) 2.5 times as long Answer: c) Twice as long

Step-by-step solution

Identify the known information and the question

We are given that Adam's initial speed is half of Bob's initial speed: let \(v_{A0} = 0.5v_{B0}\), where \(v_{A0}\) is Adam's initial speed and \(v_{B0}\) is Bob's initial speed. We need to compare the time they spend in the air.

Understand how time in the air is related to initial velocity

When an object is thrown upwards, it reaches its maximum height when its vertical velocity becomes 0 (momentarily) before falling back down. To find the time it took to reach that height, we can use the kinematic equation: $$v = v_{0} - gt$$ where \(t\) is the time to reach the maximum height, \(v\) is the final velocity (0 m/s at max height), \(v_{0}\) is the initial velocity, and \(g\) is the gravitational acceleration (9.8 m/s²).

Calculate the time each person spends in the air

First, we'll solve the kinematic equation for time for both Adam and Bob separately. For Adam: $$0 = v_{A0} - gt_{A}$$ $$t_{A} = \frac{v_{A0}}{g} = \frac{0.5v_{B0}}{g}$$ For Bob: $$0 = v_{B0} - gt_{B}$$ $$t_{B} = \frac{v_{B0}}{g}$$

Compare the time each person spends in the air

Now, we'll find the ratio between the times each person spends in the air. Divide the equation for \(t_A\) by the equation for \(t_B\): $$\frac{t_A}{t_B} = \frac{\frac{0.5v_{B0}}{g}}{\frac{v_{B0}}{g}}$$ Simplify the expression for the ratio: $$\frac{t_A}{t_B} = \frac{0.5v_{B0}}{v_{B0}}$$

Identify the correct answer

For the ratio between the times each person spends in the air: $$\frac{t_A}{t_B} = 0.5$$ This means that Bob spends twice as long in the air as Adam, so the answer is c) twice as long.