A 55 kg human cannonball is shot out the mouth of a 4.5 m cannon with a speed of

Question

A 55 kg human cannonball is shot out the mouth of a 4.5 m cannon with a speed of 18 m/s at an angle of 60*. Friction and air resistance are negligible in this problem. You may not use Newton's laws or the equations of motion to solve these problems. (Think conservation of energy). Determine the following quantities for the human cannonball as she exits the mouth of the cannon.

a. the horizontal and vertical components of her velocity

b. her kinetic energy

Determine the following quantities for the human cannonball at the top of her trajectory.

c. the horizontal and vertical components of her velocity

d. her kinetic energy

e. her potential energy relative to the mouth of the cannon

f. her height above the mouth of the cannon

Explanation / Answer

Given,

m = 55 kg ; v0 = 18 m/s ; L = 4.5 m ; theta = 60 deg

a)v0x = 18 cos60 = 9 m/s

v0y = 18 sin60 = 15.6 m/s

Hence, v0x = 9 m/s ; v0y = 15.6 m/s

b)KE = 1/2 m v^2

KE = 0.5 x 55 x 18^2 = 8910 J

Hence, KE = 8910 J

c)v0x remains same but v0y becomes zero

v0x = 9 m/s ; v0y = 0 m/s

d)KE = 0 J at the top

e)PE = KE(inital) as no non conservative forces are acting on her.

PE = 8910 J

f)m g h = 8910 J

h = 8910/55 x 9.8 = 16.53 m

Hence, h = 16.53 m

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