Final answer:
The velocity of a person just before their feet strike the ground, having jumped from a 3.7-m high house roof, is calculated using the kinematic equation for free fall. After performing the calculation, the final velocity is found to be approximately 8.52 m/s to two significant figures.
Step-by-step explanation:
To find the velocity of a person just before his feet strike the ground after jumping from the roof of a house 3.7-m high, we can use the equations of motion under constant acceleration due to gravity. Ignoring air resistance, the velocity can be calculated using the following kinematic equation for free fall:
v^2 = u^2 + 2g*h
Where:
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- v is the final velocity
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- u is the initial velocity (which is 0 m/s when starting from rest)
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- g is the acceleration due to gravity (9.81 m/s^2)
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- h is the height the person falls from (3.7 m)
Substituting the known values into the equation, we get:
v^2 = 0^2 + 2 * 9.81 m/s^2 * 3.7 m
Calculating the final velocity, we have:
v = √(2 * 9.81 m/s^2 * 3.7 m)
v = √(72.582 m^2/s^2)
v = 8.52 m/s (to two significant figures)
Therefore, the velocity of the person's torso, with a mass of his torso (excluding legs) is 46 kg, just before his feet strike the ground is approximately 8.52 m/s when rounded to two significant figures.