Posted by **Jean** on Thursday, March 15, 2007 at 5:38am.

A 20.0 kg cannon ball is fired from a cannnon with muzzle speed of 1000 m/s at an angle of 37.0 with the horizontal. A second ball is fired at an angle of 90.0. Use the conservation of energy principle to find

a] the maximum height reached by each ball and

b] the total mechanical energy at the maximum height for each ball. Let y=0 at the cannon

For the first cannon, find the vertical component of velocity. Because the horizontal velocity is constant, the portion of KE that converts to gpe is given by 1/2 m vvertical^2.

mgh= 1/2 m vvertical^2 Solve for h.

a) Let Voy be the INITIAL vertical velocity component of either cannonball.

The maximum height H reached is given by

g H = (1/2)Voy^2

For the cannonball fired at 37 degrees,

Voy = 1000 sin 37 = 601.8 m/s

for the one fired vertially,

Voy = 1000 m/s

Now complete the calculation of h for each case

b) You can save your self severalsteps by using the fact that the total mechanical energy remains equal to the initial kinetic energy:

Etotal = (1/2) M V^2

- physics. -
**Amal Alzamil**, Friday, November 25, 2011 at 6:31am
A 20.0-kg cannon ball is fired from a cannon with a muzzle speed of 1000 m/s at an angle of 37.0° with the

horizontal. A second ball is fired at an angle of 90.0°. Use the conservation of energy principle to find

(a) the maximum height reached by each ball and

For the first cannon ball:

vy = 1000 sin 37 = 601.8 m/s

K = ½ m v² = U = m g h h = v² / 2g = (601.8)² / 2 / 9.8 = 18,478 m

For the second cannon ball:

vy = 1000 sin 90 = 1000 m/s

K = ½ m v² = U = m g h h = v² / 2g = (1000)² / 2 / 9.8 = 51,020 m

(b) the total mechanical energy at the maximum height for each ball. Let y = 0 at the cannon.

Total energy = ½ m v² = (0.5) (20) (1000)² = 10

7

J

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