A 22.0-kg cannonball is fired from a cannon with muzzle speed of 950 m/s at an a

Question

A 22.0-kg cannonball is fired from a cannon with muzzle speed of 950 m/s at an angle of 34.0 degree with the horizontal. A second ball is fired with the same initial speed at an angle of 90.0 degree. Let y = 0 at the cannon. (a) Use the isolated system model to find the maximum height reached by each ball. h_First ball = m h_second ball = m (b) Use the isolated system model to find the total mechanical energy of the ball-Earth system at the maximum height for each ball. E_first ball = J E_second ball = J A bead slides without friction around a loop-the-loop (see figure below). The bead is released from rest at a height h = 3.10R. (a) What is its speed at point ? (Use the following as necessary: the acceleration due to gravity g, and R.)

Explanation / Answer

m = 22.0 kg, v = 950 m/s, = 34o

(a) 1st Ball, vy = 950 * sin34 = 531.23 m/s

vf^2 = vy^2 - 2*g*s
0 = 531.23^2 - 2*9.8*s
s = 14398.23 m
Max height reached by 1st ball, s = 14398.23 m

2nd Ball,
vy = 950 m/s
vf^2 = vy^2 - 2*g*s
0 = 950^2 - 2*9.8*s
s = 46045.91 m
Max height reached by 2nd ball, s = 46045.91 m

(b)
For 1st Ball = m*g*h = 22.0 * 9.8 * 14398.23
U = 3104258.38 J

For 2nd Ball = m*g*h = 22.0 * 9.8 * 46045.91
U = 9927500 J

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.