boy of mass 64.1 kg is rescued from a hotel fire by leaping into a firefighters' net. The window from which he leapt was 7 m above the net. The firefighters lower their arms as he lands in the net so that he is brought to a complete stop in a time of 0.40 s. Ignore air resistance.

(a) What is his change in momentum during this time interval?

(b) What is the impulse on the net due to the boy during the interval? [Hint: Do not ignore gravity.]

(c) What is the average force on the net due to the boy during the interval?

PE =m g h= 64.1 * 9.81 * 7 = 4402 oules

so
(1/2) m v^2 = 4402
32.05 v^2 =4402
v = 11.72 meters/second
P = m v = 64.1 * 11.72 = 751 kg m/s
impulse = force * time =change in momentum = 751 up on boy
so
F = 751 / 0.4 ---> 1878 Newtons downward on net

what is difference between force in impulse formula and final force

To find the answers to these questions, we can use the principles of momentum and impulse.

(a) To calculate the change in momentum (Δp), we can use the formula:

Δp = m * Δv

Where m is the mass of the boy and Δv is the change in velocity.

Since the boy is brought to a complete stop, his initial velocity (vi) is the velocity at which he jumped out of the window, and his final velocity (vf) is 0 m/s.

The change in velocity (Δv) is then vf - vi = 0 - vi = -vi.

Therefore, the change in momentum is:

Δp = m * Δv = m * (-vi)

Substituting the given values:

Δp = 64.1 kg * (-vi)

(b) Impulse (J) is defined as the change in momentum, so:

J = Δp

Using the result from part (a), we find:

J = 64.1 kg * (-vi)

(c) Average force (F) during an interval of time is given by the formula:

F = J / Δt

Where Δt is the time interval.

Substituting the given values, we have:

F = J / 0.40 s

Now let's calculate the values:

(a) Δp = -64.1 kg * vi

(b) J = -64.1 kg * vi

(c) F = (-64.1 kg * vi) / 0.40 s