Inelastic Human Cannonball

Here is the problem:

In a circus act, Marcello (mass 70kg) is shot from a cannon with a muzzle velocity of 24.00 m/s at an angle of 30 degrees above the horizontal. His partner, Tina (mass 50kg), stands on an elevated platform located at the top of his trajectory. He grabs her as he flies by and the two fly off together. They land in a net at the same elevation as the cannon a horizontal distance x away. Find x.

The text gives a partial solution to this one. However, I cannot figure out how to get the final velocity, unless it is equated somehow with the initial velocity.

Vxi = (24cos30) m/s = (70*20.8)kgm/s = (120* Vxf)kgm/s;

Vxf = 12.1 m/s

time of flight to the platform: t = (24 sin 30)/gs = 1.22s

x = (20.8 + 12.1)1.22m = 40.1 m

The part I do not understand, is how to find the final velocity.

Thanks for your help!

I think that throughout our investigation we need to assume that the friction of their passage through the air is negligible. Otherwise the problem is not solvable with any reasonable effort. We can not however, assume that all friction is negligible, otherwise he will never be able to hold onto her and exert sufficient force to bring her velocity from zero up to their common post-grabbing velocity. Clearly there must be significant friction between Marcello and Tina so that they can continue the flight as one. The presence of this necessary friction in the system means that the conservation of mechanical energy does not hold. This throws out that handy tool of requiring that the sum of potential and kinetic energy be constant.

We do know one quantity that is conserved in an inelastic collision of the sort described here. That is momentum. We hear nothing of any rotation of the people involved in this adventure so we will assume that only linear momentum is involved. Linear momentum is to product of mass times velocity. If it is to be conserved then the sum of M and T's momentum before they connect must be equal to the sum of their momentum afterward. Before they connect M has all the momentum and at the moment of collision it is entirely horizontal since they connect at the top of the trajectory where the vertical component of velocity is zero.

In the absence of air resistance the horizontal velocity of M does not change until he reaches T. It remains 12.1 m/s. His momentum before collision then is 70*12.1=847 kgm/s. Conservation of momentum demands that immediately after they connect the combined momentum be also 847 kgm/s since T's contribution is zero. To get the final horizontal velocity we need only divide the momentum by their combined mass vf=847/120=7.06m/s.

Next we need to know the time required for them to fall back to the level of the cannon. We know that the initial vertical velocity was sufficient to require 1.22 seconds for gravitational acceleration to use it all up., bringing the vertical component to zero. It follows that 1.22 seconds will be required for gravity to return their combined mass to the initial level. Remember that their combined weight is greater, but their mass is greater also so that their acceleration remains g. So x=1.22s*7.06 m/s = 8.61 m.

In my experience students long out of school have the advantage of being in class because they want to be. The disadvantage may be that you will have trouble being flexible in how you perceive the universe. You will have had a long successful experience based on what you understand and may have trouble throwing some of that understanding away when you come up against some of the more advanced topics. A child has less trouble than an adult accepting that a photon is neither a wave nor a particle but may appear like one or the other depending on how you look at it.