A 19.0-kg cannonball is fired from a cannon with muzzle speed of 1150 m/s at an angle of 42.0° 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. hfirst ball =? m
hsecond 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.
Efirst ball = ? J
Esecond ball = ? J
here,
mass , m = 19 kg
initial speed , u = 1150 kg
theta1 = 42 degree
theta2 = 90 degree
a)
for first ball
the maximum height reached , h1 = (u1 * sin(theta1))^2 /( 2g)
h1 = (1150 * sin(42))^2 /(2* 9.81) = 3.02 * 10^4 m
for seccond ball
the maximum height reached , h2 = (u1 * sin(theta2))^2 /( 2g)
h2 = (1150 * sin(90))^2 /(2* 9.81) = 6.74 * 10^4 m
b)
for first ball,
the total mechanical energy of the ball-Earth system at the maximum height , ME1 = initial kinetic energy
ME1 = 0.5 * m * u^2 = 0.5 * 19 * 1150^2 J
ME1 = 1.26 * 10^7 J
for seccond ball,
the total mechanical energy of the ball-Earth system at the maximum height , ME2 = initial kinetic energy
ME2 = 0.5 * m * u^2 = 0.5 * 19 * 1150^2 J
ME2 = 1.26 * 10^7 J
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