Physics - Trampoline?

Question

A girl of mass M1 (60 kg) springs from a trampoline with an initial upward velocity (initial velocity = 8 m/s). At height 2 meters above the trampoline, the girl grabs a box of mass M2 (15 kg).

Assume 9.8 is the force of gravity.

The three questions:

1) What is the speed (v-before) of the girl immediately before she grabs the box? (Expressed in m/s)

2) What is the speed (v-after) of the girl immediately after she grabs the box? (Expressed in m/s)

3) What is the max height the girl reaches (this is measured with respect to the top of the trampoline)? (Expressed in meters)


Any and all help is greatly appreciated. Thanks! :)

Answer 1

Note: 9.8 is not the force, but the gravitational accelaration.
Question: what is the velocity of the box of mass M2 just before the girl grabs it? Assume it is zero.

You may use energy conservation to solve this problem.
1) the initial kinetic energy of the girl:
Ki = 0.5*M1*v^2 = 0.5*60*8^2 (J)
At height 2 meters above the trampoline, the girl's potential is:
P = M1*g*h = 60*9.8*2 (J)
K = Ki - P is her kinetic energy at that height 2 meters above the trampoline. Since 0.5M1*V^2 = K = Ki - P
You have the girl's speed at 2m high:
V = sqrt(2*K/M1) (in m/s)
where sqrt means "the square root of".

2) Again use energy conservation. Notice the total kinetic energy is still K but the total mass is now (M1+M2). thus:
V = sqrt(2*K/(M1+M2)) (in m/s)

3) Again use energy conservation. Notice the total kinetic energy is still K, and this kinetic energy would be converted to the potential energy of M1 and M2. The additional height beyond the first 2m is:
K/ ((M1+M2)*g) (in m)

Please do the math.