Verified answer

conservation of energy:

potential energy gain = loss in kinetic energy

=> mgh = Δ[½mv²]

=> gh = ½[8² - v²]

=> v² = -2(9.80)2 + 64 = 103.2

=> v = 4.98 m/s (upwards)

conservation of momentum(mv):

momentum before impact = after

=> 4.98(60) = 75v'

=> v' = 3.98 m/s (upwards)

with this velocity, maximum height (above the trampoline reached ) = 2 + 3.98²/(2g) = 2 + 3.98²/19.6 ~= 2.81 m

hope this helps

Trampoline Box

"A girl of mass m1=60.0 kilograms springs from a trampoline with an initial upward velocity of vi=8.00 meters per second."

That tells you her kinetic energy (1/2)mv^2. Calculate it.

"At height h=2.00 meters above the trampoline,"

She has lost energy mgh. Calculate it. This will tell you her new kinetic energy, which in turn will tell you the answer to the first question (her speed at this point).

"the girl grabs a box of mass m2=15.0 kilograms."

Momentum is conserved. So calculate her total momentum. Then set it equal to (m1 + m2)*v, the momentum of girl + box, and solve for her new v.

"What is the maximum height hmax that the girl (with box) reaches?"

Until her new kinetic energy (1/2)mv^2 is converted to gravitational potential energy mgh. Set them equal and solve for h. That would be the height relative to where she grabbed the box.