A boy of mass 37.8kg. is rescued from a hotel fire by leaping into a firefighters' net. The window from which he leapt was 8.02 above the net. The firefighters lower their arms and he lands in the net so that he is brought to a complete stop in a time of 0.491 s.

What is his change in momentum durin gthe 0.497-s interval.

To calculate the change in momentum, we need to know the initial momentum and final momentum of the boy.

The momentum of an object is calculated using the formula: momentum (p) = mass (m) × velocity (v).

In this scenario, we are given the mass of the boy (m = 37.8 kg) and the time it takes for him to come to a complete stop (t = 0.491 s). However, we don't have the information about his initial or final velocity.

To find the initial velocity, we can use the equation of motion: v = u + at.
Here, u represents the initial velocity, a represents acceleration, and t represents time.

In this case, we know that the boy comes to a complete stop, so his final velocity (v) is 0 m/s. The acceleration can be calculated using the equation: a = (v - u) / t.

Substituting the values:
0 m/s = u + a × t
0 m/s = u + a × 0.491 s

Since the boy comes to a stop, his final velocity is zero.

Now, we have two equations:
p_initial = m × u
p_final = m × v

As the boy comes to a complete stop, his initial momentum (p_initial) is equal to his final momentum (p_final).
So, p_initial = p_final.

This implies that m × u = m × v. Since m is common on both sides of the equation, it cancels out:

u = v

Substituting u with v = 0 m/s in the first equation, we get:

0 m/s = 0 m/s + a × 0.491 s

Simplifying the equation, we have:

0 m/s = 0.491 s × a

Since the left side of the equation is zero, a must also be zero.

Therefore, the acceleration of the boy during the stopping interval is 0 m/s². This means that there is no change in velocity during this time.

As a result, the change in momentum during the 0.491 s interval is also zero.