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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. -

    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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