Part A

Use $d=\frac{1}{2} g t^{2}$ and solve for the time one spends moving upward in a $0.6-\mathrm{m}$ vertical jump. Then double it for the "hang time" --the time one's feet are off the ground.

Express your answer with the appropriate units.
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$$
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$14: 40$
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2024/9/11

Question

Part A

Use $d=\frac{1}{2} g t^{2}$ and solve for the time one spends moving upward in a $0.6-\mathrm{m}$ vertical jump. Then double it for the "hang time" --the time one's feet are off the ground.

Express your answer with the appropriate units.
$$
t=\text { Value Units }
$$
Submit
Request Answer
Provide Feedback
Next)
$14: 40$
搜索
yd
দ
2024/9/11

Accepted Answer

∻Solution∻ ‖ a ⋮ To find the time spent moving upward in a vertical jump of 0.6 meters, we use the equation for displacement under constant acceleration due to gravity: $$d = \frac{1}{2} g t^2$$. Here, $d = 0.6 \, \mathrm{m}$ and $g = 9.8 \, \mathrm{m/s^2}$. Rearranging the equation to solve for time $t$, we have: $$t = \sqrt{\frac{2d}{g}} = \sqrt{\frac{2 imes 0.6}{9.8}}$$. Calculating this gives us the time spent moving upward. ‖ ‖ b ⋮ After calculating the time $t$, we then double this value to find the total "hang time," which is the total time the person's feet are off the ground. The hang time is given by: $$ ext{Hang time} = 2t$$. We will express the final answer in seconds. ‖ ∻Answer∻ ⚹ [Insert final answer here] ⚹ ∻Key Concept∻ ⚹ Kinematics of free fall: The displacement of an object under the influence of gravity can be described using the equation $$d = \frac{1}{2} g t^2$$, where $d$ is the displacement, $g$ is the acceleration due to gravity, and $t$ is the time. ⚹ ∻Explanation∻ ⚹ The problem involves calculating the time of a vertical jump using kinematic equations, which allows us to determine both the time spent moving upward and the total hang time. ⚹

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