A 31 kg cannon ball is fired from a cannon with muzzle speed of 864 m/s at an angle of 32.9 degrees?

Question

A 31 kg cannon ball is fired from a cannon with muzzle speed of 864 m/s at an angle of 32.9 degrees with the horizontal.
Gravity is 9.8 m/s/s.
A.) Use conservation of mechanical energy to find the maximum height reached by the ball.

B.) A second identical ball is fired at an angle of 132 degrees.
What is teh total mechanical energy at the maximum height of the second ball. Answer in units of J.

Please show your work and thanks for the help!
Happy Holidays!!!!!

Answer 1

TE = KE; the total energy at the muzzle is kinetic energy of which KE = KE(v) + KE(h) kinetic energy vertical and horizontal. Of these two KE(v) = mgH = PE(H); where m = 31 kg, g = 9.81 m/sec^2, and H is the max height above the muzzle, the answer you are looking for.

A) Thus KE(v) = 1/2 mvy^2 = mgH and H = vy^2/2g; where vy = v sin(theta) and v = 864 mps and theta = 32.9 degrees. You can do the math.

B) KE = KE(v) + KE(h); all the KE(v) is converted into PE(H); so only the KE(h), the horizontal kinetic energy, remains when the ball reaches H. Thus, the mechanical energy at H is KE(h) = 1/2 mvx^2; where vx = v cos(theta) and theta now = 132 degrees. You can do the math.

The physics is this. The vertical kinetic energy is converted by the force of gravity into potential energy, while the horizontal kinetic energy remains as kinetic energy unless something, like drag friction, changes it.

The Joule is an arcane term; it is better to keep the answers in kg m/sec^2, which are the SI units for a Joule. The SI units are far more useful and meaningful than Joule.