Answer
(a) The boy's change in momentum is $-751.2~kg~m/s$
(b) The impulse on the net is $986.4~kg~m/s$
(c) The average force on the net due to the boy is $2466~N$
Work Step by Step
(a) We can find the speed after falling 8.0 meters:
$v_f^2 = v_0^2+2ay$
$v_f = \sqrt{v_0^2+2ay}$
$v_f = \sqrt{0+(2)(9.80~m/s^2)(8.0~m)}$
$v_f = 12.52~m/s$
We can find the change in momentum:
$\Delta p =m~\Delta v = (60.0~kg)(-12.52~m/s) = -751.2~kg~m/s$
The boy's change in momentum is $-751.2~kg~m/s$
(b) The impulse on the net is the sum of the boy's initial momentum and the gravitational force during the collision:
$J = p_0+mg~t$
$J = (751.2~kg~m/s)+(60.0~kg)(9.80~m/s^2)(0.40~s)$
$J = 986.4~kg~m/s$
The impulse on the net is $986.4~kg~m/s$
(c) We can find the average force on the net due to the boy:
$F~t = 986.4~kg~m/s$
$F = \frac{986.4~kg~m/s}{t}$
$F = \frac{986.4~kg~m/s}{0.40~s}$
$F = 2466~N$
The average force on the net due to the boy is $2466~N$
