A medieval city has the shape of a square and is protected by walls with length 550 m and height 16 m. You are the commander of an attacking army and the closest you can get to the wall is 100 m. Your plan is to set fire to the city by catapulting heated rocks over the wall (with an initial speed of 80 m/s). At what range of angles should you tell your men to set the catapult? (Assume the path of the rocks is perpendicular to the wall. Round the answer to one decimal place. g ≈ 9.8 m/s2)
I have no idea where to start.
To determine the range of angles that should be used for the catapult, we can start by analyzing the motion of the heated rocks. The key is to find the angle that will maximize the distance traveled by the rocks while still clearing the 16-meter-high wall.
Let's break down the problem into components:
1. Horizontal Motion: The horizontal velocity of the rocks is 80 m/s and remains constant throughout the motion due to the absence of any horizontal forces. Therefore, the time it takes for the rock to travel a certain horizontal distance is the same for any angle.
2. Vertical Motion: The vertical distance the rocks need to clear is 16 m. We can use the equation of motion in the vertical direction:
Δy = v₀y * t + (1/2) * g * t^2
where:
- Δy is the vertical displacement (16 m).
- v₀y is the initial vertical velocity (unknown).
- t is the time of flight (unknown).
- g is the acceleration due to gravity (9.8 m/s^2).
Rearranging the equation, we have:
t = (√(2 * Δy / g))
Since the time of flight is the same for both the upward and downward portions of the motion, the total time is t = 2 * (√(2 * Δy / g)).
3. Maximum Range: The maximum horizontal distance the rocks can travel is achieved when the rocks hit the ground on the other side of the wall. At that point, the vertical displacement Δy is equal to the height of the wall (16 m).
Putting all the information together:
Range = v₀x * t
= v₀ * cos(θ) * (2 * (√(2 * Δy / g)))
Now, to determine the range of angles, we need to find the maximum value for the range equation.
To maximize the range, we can differentiate the equation with respect to θ, set it equal to zero, and solve for θ:
d(Range)/d(θ) = -v₀ * sin(θ) * (2 * (√(2 * Δy / g))) + v₀^2 * cos(θ) * 2 * (1/2) * (√(2 * Δy / g)^-3) = 0
Simplifying the equation:
-sin(θ) * (√(2 * Δy / g)) + v₀ * cos(θ) / (√(2 * Δy / g)) = 0
Dividing both sides by (√(2 * Δy / g)) and rearranging, we have:
tan(θ) = v₀ / (√(2 * Δy * g))
Now, we can solve for θ by taking the inverse tangent of both sides:
θ = arctan(v₀ / (√(2 * Δy * g)))
Plugging in the given values:
θ = arctan(80 / (√(2 * 16 * 9.8)))
Using a calculator, we can find:
θ ≈ 48.7 degrees (rounded to one decimal place)
Therefore, the range of angles you should tell your men to set the catapult is between approximately 0 and 48.7 degrees (inclusive), ensuring the heated rocks clear the 16-meter-high wall while maximizing the distance traveled.
To solve this problem, we first need to understand the motion of the projectile (heated rocks) when launched by the catapult. We can break down the motion into horizontal and vertical components.
The horizontal motion is constant, as there is no horizontal force acting on the projectile once it is launched. The vertical motion is affected by the force of gravity. Using these components, we can determine the time of flight and the range of the projectile.
Let's begin by finding the time of flight. We can use the vertical motion equation to find the time it takes for the projectile to reach the maximum height:
h = (v^2*sin^2θ) / (2g),
where h is the maximum height reached, v is the initial velocity, θ is the launch angle, and g is the acceleration due to gravity.
In this case, the maximum height is equal to the height of the city wall, 16 m. The initial velocity is given as 80 m/s, and the acceleration due to gravity is approximately 9.8 m/s^2.
16 = (80^2*sin^2θ) / (2*9.8).
Next, we can solve this equation for sin^2θ:
sin^2θ = (16 * 2 * 9.8) / (80^2).
sin^2θ ≈ 0.098.
Taking the square root of both sides, we find:
sinθ ≈ 0.313.
Now, let's find the launch angle θ. To do this, we can use the inverse sine function:
θ ≈ arcsin(0.313).
θ ≈ 18.1 degrees.
Note that sinθ has two positive solutions, one in the first quadrant and another in the second quadrant. However, in this case, we can disregard the second quadrant angle since it would cause the rocks to be launched backward, away from the city wall.
Finally, to calculate the range of angles you should tell your men to set the catapult, we need to consider the range of values for θ. Since the path of the rocks is perpendicular to the wall, the angles should be between 0 and 90 degrees. Therefore, the range of angles should be approximately 0 to 18.1 degrees.
In summary, you should instruct your men to set the catapult at angles between 0 and 18.1 degrees to hit the target effectively.
the range R of a projectile fired with angle v at angle θ is
R = v^2 sin(2θ)/g
You have the numbers; solve for θ and you should get two answers. Those will be the minimum and maximum θ which for that range.