Look at the question posted by newtonn on jan 18 page 21 and the answer by nessus.

Posted on

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

Posted on

## We’ll help your grades soar

Start your 48-hour free trial and unlock all the summaries, Q&A, and analyses you need to get better grades now.

- 30,000+ book summaries
- 20% study tools discount
- Ad-free content
- PDF downloads
- 300,000+ answers
- 5-star customer support

Already a member? Log in here.

Are you a teacher? Sign up now