3) A Human cannonball (m=68 kg) is loaded into a spring loaded canon angled at 4

Question

3) A Human cannonball (m=68 kg) is loaded into a spring loaded canon angled at 45 degrees above the horizontal, and is to be fired across a stadium. The spring is compressed 1.5 meters from the muzzle of the cannon. Once the cannon is fired it takes 0.13 s for the human cannonball to exit the front of the cannon. SHOW ALL WORK

A) If the Impulse inside the cannon is 1388 N*s what is the muzzle velocity of the cannon?

B) How far away should the airbag be placed from the front of the cannon so the human cannonball isn’t hurt?

Explanation / Answer

here,

mass , m = 68 kg

theta = 45 degree

compression in the spring , x = 1.5 m

time taken , t = 0.13 s

(a)

the impulse , I = 1388 N.s

average force exerted , Fvg = I/t

Favg = 10676.92 N

the average accelration of man , a = Favg/m

a = 157.01 m/s^2

if x = 1.5 m

let the muzzle velocity be v

using third equation of motion

v^2 - u^2 = 2*a*x

v^2 = 2*157.01 * 1.5

v = 21.7 m/s

the muzzle velocity of the cannon is 21.7 m/s

(b)

the range , R = ( v^2 * sin(2*theta) )/g

R = 21.7^2 *sin( 2*45) / 9.8

R = 48.07 m

the airbag be placed at 48.07 m from the front of the cannon

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