With what speed does the mass hit the ground at the bottom of the cliff?
The following takes place at a colony on the Moon some time in the distant future:
A 40 kg cannon is loaded with a 10 kg cannonball and mounted on a 6 kg platform with wheels very low to the ground. The wheeled platform is pushed and moves forward at 3 m/s horizontally. When the cannonball is fired out, the cannon recoils (backward) with a velocity of 2 m/s relative to the ground. The cannonball then strikes a large solid block that weighs 32N and is sitting on the edge off a 50 m high cliff and becomes embedded in it. The combined mass is shot horizontally off the cliff. With what speed does the mass hit the ground at the bottom of the cliff?
Look at the question posted by newtonn on jan 18 page 21 and the answer by nessus.
All the information about the cannon and its forward speed and recoil speed are extra information, not necessary to solve the problem. It also does not matter what the mass of the rock is, with or without the cannon ball embedded in it.
The key is to remember that horizontal and vertical velocities are treated as acting independently.
Thus, when the cannon ball hits the rock and pushes it over the cliff its initial velocity in the downward or y-direction is zero. The question is, how much will its velocity increase as it falls the 50 meters under the influence of the moon's gravitational acceleration. The value for g on the moon is 1.63 m/s/s.
Now using a kinematic equation: Vf^2 = Vi^2 +2gh, you can solve for the final velocity (Vf)
Vf^2 = 0 + 2 x 1.63 x 50
and Vf = 12.77 m/s
i dont think the person answered right. can someone else help me?