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A 60kg resident jumps from the first floor of a burning house. His velocity just before landing on the ground is 6 ms-¹.

(a) calculate the impulse when his legs hit the ground​

User Vi
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Answer:

The impulse when his legs hit the ground is approximately 359 N s

Explanation:

To calculate the impulse, we need to use the formula:

Impulse = change in momentum

Since the resident jumps from rest, his initial momentum is zero. We can calculate his final momentum using the formula:

p = m*v

where p is momentum, m is mass, and v is velocity.

p = 60 kg * 6 m/s = 360 kg m/s

Therefore, the change in momentum is:

Δp = final momentum - initial momentum = 360 kg m/s - 0 = 360 kg m/s

The impulse is equal to the change in momentum:

Impulse = Δp = 360 kg m/s

However, the question specifically asks for the impulse when his legs hit the ground. This means we need to consider the time it takes for his legs to come to a stop after hitting the ground. The impulse is given by:

Impulse = force x time

We can rearrange this formula to solve for time:

time = Impulse / force

To find the force, we can use the formula:

force = mass x acceleration

The resident is brought to a stop by the ground, so we can assume that the force exerted by the ground is equal to the resident's weight, which is:

force = mass x gravity

where gravity is the acceleration due to gravity, approximately 9.81 m/s².

force = 60 kg x 9.81 m/s² = 588.6 N

Now we can calculate the time it takes for the resident's legs to come to a stop after hitting the ground:

time = Impulse / force = 360 kg m/s / 588.6 N ≈ 0.61 s

Therefore, the impulse when his legs hit the ground is:

Impulse = force x time ≈ 588.6 N x 0.61 s ≈ 359 N s

So the impulse when his legs hit the ground is approximately 359 N s.
User Kocus
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