Question

An arrow is shot horizontally with a speed of \(20 \mathrm{~m} / \mathrm{s}\) from the top of a tower \(60 \mathrm{~m}\) high. The time to reach the ground will be a) \(8.9 \mathrm{~s}\) c) \(3.5 \mathrm{~s}\) e) \(1.0 \mathrm{~s}\) b) \(7.1 \mathrm{~s}\) d) \(2.6 \mathrm{~s}\)

Short answer

a) 1.5 s b) 3.0 s c) 3.5 s d) 4.0 s Answer: c) 3.5 s

Step-by-step solution

Identify the given values and what needs to be found

The problem provides the initial horizontal velocity (\(20 \mathrm{~m} / \mathrm{s}\)), the height of the tower (\(60 \mathrm{~m}\)), and the goal is to find the time it takes for the arrow to reach the ground.

Horizontal and Vertical Motion

Since the arrow's motion can be broken down into horizontal and vertical components, it's important to recognize that the horizontal velocity stays constant throughout the motion, while the vertical velocity and position will change due to the force of gravity.

Use the vertical motion to find the time

The equation for the vertical position (\(y\)) of an object under constant acceleration (in this case, gravity) can be written as: \(y = y_0 + v_{0y}t + \frac{1}{2}at^2\) Where \(y\) is the final vertical position, \(y_0\) is the initial vertical position, \(v_{0y}\) is the initial vertical velocity, \(t\) is the time, and \(a\) is the vertical acceleration due to gravity (\(-9.81 \mathrm{~m} / \mathrm{s}^2\)). In this case, since the arrow is shot horizontally, the initial vertical velocity \(v_{0y}\) is \(0\). The initial vertical position \(y_0\) is the height of the tower (\(60 \mathrm{~m}\)), and since the arrow falls to the ground, the final vertical position \(y\) is \(0\). We can rewrite the equation as: \(0 = 60 + \frac{1}{2}(-9.81)t^2\)

Solve for time

To find the time it takes for the arrow to reach the ground, we can solve the equation for \(t\). First, isolate the term with time: \(t^2 = \frac{2\cdot 60}{9.81}\) Now, take the square root of both sides to solve for time: \(t = \sqrt{\frac{2\cdot 60}{9.81}} \approx 3.5 \mathrm{~s}\)

Check the answer and provide the solution

The value of \(t\) matches one of the available answer choices (option c). Therefore, the time it takes for the arrow to reach the ground is approximately \(3.5 \mathrm{~s}\).

Key concepts

Horizontal Projectile Motion

Horizontal projectile motion is an intriguing concept where an object moves in a two-dimensional plane with a constant horizontal velocity and an acceleration only in the vertical direction due to gravity. Imagine you are throwing a stone off a cliff; as the stone travels forward, it also starts descending, tracing a curved path through the air. This happens because, in the absence of air resistance, the horizontal velocity remains unchanged throughout the motion, while the vertical motion is influenced by gravity.

In the example of the arrow shot from the tower, the arrow's initial horizontal speed (\(20 \text{ m/s}\)) won't change as it falls to the ground. However, its vertical speed increases due to gravity, making the downward journey accelerate over time. By understanding horizontal projectile motion, students can predict not just how long an object will stay in the air but also how far it will travel before hitting the ground.

Free Fall Acceleration

Free fall acceleration is crucial to understanding the vertical component of projectile motion. It is the acceleration experienced by an object solely under the influence of gravity, without any other forces acting on it, such as air resistance. On Earth, the standard acceleration due to gravity is approximately \(9.81 \text{ m/s}^2\) downwards.

When we say an object is in free fall, we imply that gravity is the sole force acting on it. For the arrow in the exercise, once it leaves the bow, it is subject to free fall acceleration in the vertical direction. This principle allows us to predict the time it takes for the arrow to reach the ground by only considering its interaction with gravity.

Kinematic Equations

The kinematic equations are the tools that tie together velocity, acceleration, and position of an object in motion, particularly when that motion has a constant acceleration like free fall. There are four key kinematic equations, but the one relevant to our arrow example is:\[ y = y_0 + v_{0y}t + \frac{1}{2}at^2 \]
where \(y\) is the final position, \(y_0\) is the initial position, \(v_{0y}\) is the initial velocity in the y-direction, \(t\) is time, and \(a\) is the constant acceleration (in this case due to gravity).

Using this kinematic equation, we could derive the time taken for the arrow to fall to the ground considering its initial vertical velocity \(v_{0y}\) is 0 since the arrow was shot horizontally. Kinematic equations simplify complex motion into manageable calculations to understand the dynamics of moving objects better.

Constant Acceleration

Constant acceleration is a uniform acceleration, meaning that the velocity of an object changes at a consistent rate over time. This concept forms the foundation for the kinematic equations used in physics to analyze motion. When dealing with projectile motion, we often assume acceleration due to gravity is constant. It's this assumption that simplifies the arrow's descent into a problem solvable by basic algebra.

In our scenario, the arrow experiences constant acceleration in the vertical direction only, and it's this acceleration that determines the time it takes for the arrow to hit the ground. The constant nature of gravitational acceleration allows students to predict the time of flight without having to consider complex forces or varying accelerations.