Answer:
571.04 N
Step-by-step 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
Answer:
Magnitude of force = 6387.39N
Step-by-step 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
