A 16.0-kg cannonball is fired from a cannon with muzzle speed of 1 050 m/s at an

Question

A 16.0-kg cannonball is fired from a cannon with muzzle speed of 1 050 m/s at an angle of 40.0° with the horizontal. A second ball is fired with the same initial speed at an angle of 90.0°. 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

Explanation / Answer

a) Verticakl velocity of the first cannonball is

vy (first ball) = vo sin40o = (1050 m/s)sin40o = 674.93 m/s

The maximum height reach by the first ball

h(first ball) = [( vy (first ball) )2 / (2 g)] { g=9.81 m/s2 }

h(first ball) =[ (674.93 m/s)2 / (2 x 9.81 m/s2 )]

h(first ball) = 23217.66 m.

Verticakl velocity of the second cannonball is

vy (second ball) = vo sin40o = (1050 m/s)sin90o = 1050 m/s

The maximum height reach by the second ball

h(second ball) = [( vy (second ball) )2 / (2 g)] { g=9.81 m/s2 }

h(second ball) =[ (1050 m/s)2 / (2 x 9.81 m/s2 )]

h(second ball) = 56192.66 m.

Answer : h(first ball) = 23217.66 m.

h(second ball) = 56192.66 m.

b) The mechanical energy at the maximum height for first ball

E(first ball) = m g h(first ball) + (1/2) m vx2 { vx =horizontal velocity of the ball =(1050 m/s)cos400 =804.35m/s}

E(first ball) = (16.0 kg)(9.81 m/s2 )(23217.66 m) + (1/2)((16.0 kg)(804.35m/s)2

E(first ball) = 3644243.91 J + 5175831.38 J= 8820075.29 J

E(first ball) =8820075.29 J =8.82 MJ

The mechanical energy at the maximum height for second ball

E(second ball) = m g h(second ball)

E(second ball) = (16.0 kg)(9.81 m/s2 )(56192.66 m)

E(second ball) = 8819999.91 J = 8.82 MJ.

Answer: E(first ball) =8820075.29 J

E(second ball) = 8819999.91 J

The mechanical energy for both balls is approximately same because both the balls are fired with same intial speed.

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