Question

The Zacchini family was renowned for their humancannonball act in which a family member was shot from a cannon using either elastic bands or compressed air. In one version of the act, Emanuel Zacchini was shot over three Ferris wheels to land in a net at the same height as the open end of the cannon and at a range of 71 m. He was propelled inside the barrel for 5.4 m and launched at an angle of 52°. If his mass was 86 kg and he underwent constant acceleration inside the barrel, what was the magnitude of the force propelling him? (Hint: Treat the launch as though it were along a ramp at 52°. Neglect air drag.)

Answer 1

Magnitude of force = 6387.39N

Explanation:

Given:

The range of motion =71m

Mass =86kg

The mass is propelled inside the barrel for 5.4m at 52°

Using projectile motion

R = Vo^2 Sin(2theta)/g

Vo^2= Rg/ sin2 theta = sqrt71×9.8/sin104

Vo= sqrt 6958/0.9703

Vo= sqrt 717.098 = 26.78m/s

Speed can be calculated using its components

Vx = Vocos theta = 26.78 cos52= 16.49m/s

Vy= Vosin theta= 26.78× sin52= 21.10m/s

Acceleration is constant

V^2 = Vo^2 + 2a◇x

a= V^2 - Vo^2/2◇x

Mass started from rest,Vo=0displacement inside the barrel will be

ax= Vx^2/2◇x = 16.49^2/(2×5.4cos52)

ax = 271.92/6.65 = 40.89m/s^2

ay= V^2/2◇y= 21.10^2/(3×5.4×sin52)

ay= 445.21/8.51 = 52.32m/s^2

The force components are

Fx=max

Fy=may + mg

Force as a vector F= Fxi + Fyj

F= (86×40.89)I + ((86×(52.32+9.8))

F= 3516.54i + 5332.59j

Magnitude of force= sqrt Fx^2+Fy^2

Magnitude of force = sqrt 12362256+ 28436516

Magnitude of force = sqrt40798772.11

Magnitude of force = 6387.39N

Answer 2

571.04 N

Explanation:

First we calculate the velocity with which he is ejected from the cannon. Using the formula for the range ,R of a projectile, R = U²sin2θ/g

So, U = √(gR/sin2θ)

g = 9.8 m/s², R = 71 m, θ = 52°

U = √(9.8 × 71 ÷ sin(2 × 52°) ) = √(695.8/sin104°) = √(695.8/0.9703) = √717.101 = 26.779 m/s ≅ 26.78 m/s

Since Emanuel Zacchini was shot from rest in the barrel of length 5.4 m, we calculate his acceleration in the barrel using v² = u² + 2as. Since u = 0, v = 26.78 m/s and s = 5.4 m, a = (v² - u²)/2s = (26.78² - 0)/(2× 5.4) = 717.1684/108 = 6.64 m/s². The force on him for this acceleration is given by F = ma , m = 86 kg, a = 6.64 m/s². So, F = 86 × 6.64 = 571.04 N ≅ 571 N