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| Math Central | Quandaries & Queries |
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Subject: physics
A person jumps from the roof of a house 3.9m high. When he strikes the ground below, he bends his knees so that the torso decelerates over an approximate distance of 0.70m. If the mass of his torso (excluding legs) is 42kg, find a) his velocity just before his feet strike the ground , and b) the average force exerted on his torso by his legs during deceleration. Please provide calculations and answers for comparison. |
Hi Jessie.
The question expects you to assume his legs are massless - not exactly realistic, but still a good question to learn from.
When the person is on the roof, his vertical speed is zero, but he quickly accelerates due to gravity until he touches the ground. The formula for this is v2 = v02 + 2ad. You know a = g and d = 3.9 m and v0 is the starting velocity which is zero m/s, so you can calculate the speed which is the answer to part (a).
For part (b), he is decelerating over a distance of 0.7 m from the speed that is the answer to part (a) down to zero. You need to know the deceleration. Again, you can use the same formula, but this time v0 is the answer from part (a), v is zero, and d is 0.7 m. Once you calculate this acceleration, you can use it in the F= ma equation to find the average force during the deceleration.
Hope this helps,
Stephen La Rocque.
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