Question

Picture yourself in the castle of Helm's Deep from the Lord of the Rings. You are on top of the castle wall and are dropping rocks on assorted monsters that are \(18.35 \mathrm{~m}\) below you. Just when you release a rock, an archer located exactly below you shoots an arrow straight up toward you with an initial velocity of \(47.4 \mathrm{~m} / \mathrm{s}\). The arrow hits the rock in midair. How long after you release the rock does this happen?

Short answer

Answer: It takes approximately 0.387 seconds for the arrow to hit the rock.

Step-by-step solution

Write down the known parameters

The height of the wall is \(18.35 \mathrm{~m}\). The initial velocity of the arrow is \(47.4\mathrm{~m}/\mathrm{s}\) upwards. We can assume that the initial velocity of the rock is \(0\mathrm{~m}/\mathrm{s}\) since it is dropped from a stationary position. The gravitational acceleration is approximately \(g = 9.81\mathrm{~m}/\mathrm{s^2}\) acting downwards.

Find equations for positions of rock and arrow

We will use the following equation of motion: $$x(t) = x_0 + v_0t - \frac{1}{2}gt^2$$ For the rock, \(x_0 = 18.35\mathrm{~m}\), \(v_0 = 0\mathrm{~m}/\mathrm{s}\), and \(g\) is acting downwards. Therefore, the equation for the rock's position is: $$x_{rock}(t) = 18.35 - \frac{1}{2}(9.81)t^2$$ For the arrow, \(x_0 = 0\mathrm{~m}\), \(v_0 = 47.4\mathrm{~m}/\mathrm{s}\) (acting upwards against the gravitational acceleration), and \(g\) is acting downwards. Therefore, the equation for the arrow's position is: $$x_{arrow}(t) = 47.4t - \frac{1}{2}(9.81)t^2$$

Find the time when the rock and arrow collide

At the collision point, both the rock and the arrow have the same position. Therefore, $$x_{rock}(t) = x_{arrow}(t)$$ $$18.35 - \frac{1}{2}(9.81)t^2 = 47.4t - \frac{1}{2}(9.81)t^2$$ Simplify the equation and solve for \(t\): $$18.35 = 47.4t$$ $$t = \frac{18.35}{47.4}$$ $$t \approx 0.387\mathrm{~s}$$

Final Answer

The arrow will hit the rock \(0.387\) seconds after the rock is released.

Key concepts

Equations of Motion

When we talk about the equations of motion, we're describing how an object moves under the influence of forces. These equations allow us to predict the future position, velocity, and acceleration of an object at any given time. In classical mechanics, the equations of motion for an object moving with constant acceleration are fundamental.

For objects moving in a straight line, the key equations are:

• The first equation of motion: \( v = u + at \), which relates velocity (v), initial velocity (u), acceleration (a), and time (t).
• The second equation of motion: \( s = ut + \frac{1}{2}at^2 \), which calculates displacement (s) based on initial velocity, acceleration, and time.
• The third equation of motion: \( v^2 = u^2 + 2as \), which connects velocity, initial velocity, acceleration, and displacement.

These equations are critical when solving kinematics problems like the one about the arrow and the rock from Helm's Deep, particularly the second equation, which was used to find the positions of the rock and arrow as functions of time.

Free Fall

Free fall is a specific type of motion where an object is subjected only to the force of gravity. In free fall, there are no other forces acting on the object, such as air resistance or friction. An important thing to remember is that all objects accelerate at the same rate when in free fall, regardless of their mass.

The acceleration due to gravity on Earth is approximately \( 9.81 \text{ m/s}^2 \) downwards. Therefore, the velocity of an object in free fall will increase by \( 9.81 \text{ m/s} \) for every second it falls, and its position will change according to the equations of motion for constant acceleration.

In the castle problem, the rock is in free fall after being dropped from the wall, which means its acceleration is equal to the acceleration due to gravity, and there is no initial upward or downward velocity.

Acceleration Due to Gravity

Acceleration due to gravity, often symbolized as \( g \), is the acceleration that Earth imparts to objects on or near its surface due to gravitational force. On Earth, \( g \) is roughly \( 9.81 \text{ m/s}^2 \) towards the center of the planet.

This value is essential when we solve problems involving objects moving under the influence of Earth's gravity, whether they're moving up or down. In our Helm's Deep scenario, both the rock and the arrow experience the same downward acceleration of \( g \), despite having different initial velocities. This is a key point: acceleration due to gravity is constant and does not depend on an object's velocity or direction of motion.

Projectile Motion

Projectile motion occurs when an object is thrown or launched into the air and moves under the influence of gravity only. This kind of motion has two components: horizontal and vertical. The horizontal motion happens at a constant velocity, while the vertical motion is subject to the acceleration due to gravity.

The arrow in our example acts as a projectile the moment it leaves the archer's bow. Even though it's initially moving straight up, it's subject to gravity, which slows it down until it stops and starts falling back down.

The rock, on the other hand, begins its motion in free fall and continues in projectile motion upon release. In both cases, we can predict the trajectory and determine the point of impact using the equations of motion for projectiles.