A bronze cannon ball with a mass of 4.5 kg is fired vertically from a small cannon and reaches a height of 10.0 m. If the canon is now aimed at an angle 33 degrees above the horizontal, calculate the horizontal distance the cannon ball would travel before it reaches the ground.
Not much of a cannon :)
time to fall to ten meters = time to rise ten meters
10 = (1/2) g t^2 = 4.9 t^2
t = 1.43 seconds up or down
v = Vi - g t
0 = Vi -9.81 (1.43)
Vi = 14 m/s muzzle velocity
now at 33 deg
u = 14 cos 33 = 11.8 m/s forever
Vi = 14 sin 33 = 7.62 initial speed up
How long to top?
0 = 7.62 - 9.81 t
t = .777 seconds going up
so
time in air = 1.55 s
so horizontal distance = 11.8*1.55 =18.3 meters
To calculate the horizontal distance the cannonball would travel before it reaches the ground, we can break down the motion into horizontal and vertical components.
First, let's calculate the time it takes for the cannonball to reach the maximum height of 10.0 m when fired vertically.
We can use the equation for the vertical displacement of an object in free fall:
Δy = V₀y * t + (1/2) * g * t²
Given:
Δy = 10.0 m (vertical displacement)
V₀y = 0 (initial vertical velocity since it was fired vertically)
g = 9.8 m/s² (acceleration due to gravity)
Rearranging the equation, we have:
10.0 m = (1/2) * (9.8 m/s²) * t²
Simplifying and solving for t:
t² = (2 * 10.0 m) / (9.8 m/s²)
t² = 2.04 s²
t ≈ √(2.04) ≈ 1.43 s
Now, let's calculate the horizontal distance the cannonball would travel when fired at an angle 33 degrees above the horizontal.
The horizontal and vertical components of the initial velocity can be calculated using trigonometry:
V₀x = V₀ * cos(θ)
V₀y = V₀ * sin(θ)
Given:
V₀ = ?
θ = 33°
To find V₀, we can use the conservation of energy. At the highest point of the cannonball's trajectory, all of its initial kinetic energy is converted to gravitational potential energy.
Kinetic Energy (KE) = Gravitational Potential Energy (PE)
KE = (1/2) * m * V₀²
PE = m * g * h
Given:
m = 4.5 kg (mass of the cannonball)
h = 10.0 m (height reached by the cannonball)
g = 9.8 m/s² (acceleration due to gravity)
Setting the two equations equal to each other:
(1/2) * m * V₀² = m * g * h
Simplifying and solving for V₀:
V₀² = 2 * g * h
V₀ = √(2 * 9.8 m/s² * 10.0 m) ≈ √(196 m²/s²) ≈ 14.0 m/s
Now, let's calculate the horizontal distance using the horizontal component of the initial velocity, V₀x.
V₀x = V₀ * cos(θ)
V₀x = 14.0 m/s * cos(33°) ≈ 11.7 m/s
To calculate the horizontal distance (D), we can use the equation of motion:
D = V₀x * t
Given:
V₀x = 11.7 m/s
t = 1.43 s
Calculating D:
D = 11.7 m/s * 1.43 s ≈ 16.7 m
Therefore, the horizontal distance the cannonball would travel before it reaches the ground when fired at an angle 33 degrees above the horizontal is approximately 16.7 meters.