Question

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. 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

Answer #1

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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