Question

In Denver, after Halloween, children bring their jack-o'-lanterns to the top of a tower and compete for accuracy in hitting a target on the ground, as shown in Figure 4.39. Suppose that the tower is \(9.0 \mathrm{~m}\) high and that the bull's-eye is a horizontal distance of \(3.5 \mathrm{~m}\) from the launch point. If the pumpkin is thrown horizontally, what is the launch speed needed to hit the bull's-eye?

Short answer

The launch speed needed is approximately \(2.59 \, \text{m/s}\).

Step-by-step solution

Understand the Problem

We have a tower with a height of \(9.0 \text{ m}\) and a target \(3.5 \text{ m}\) away horizontally. A pumpkin is thrown horizontally at a certain speed to hit the target. Our task is to find this launch speed.

Determine Time of Fall

Since the pumpkin is dropped horizontally, it falls under gravity. We use the equation of motion for vertical fall: \(h = \frac{1}{2} g t^2\), where \( h = 9.0 \, \text{m} \) (height), and \( g = 9.81 \, \text{m/s}^2 \) (acceleration due to gravity). Solving for \(t\), we have: \[ 9.0 = \frac{1}{2} \times 9.81 \times t^2. \]

Solve for Time

Let's solve the equation for time: \[ 9.0 = 4.905 t^2 \rightarrow t^2 = \frac{9.0}{4.905} \rightarrow t = \sqrt{\frac{9.0}{4.905}}. \] Upon calculation, \(t \approx 1.35 \, \text{seconds}\).

Find Horizontal Speed

Once we find the time of fall, calculate the necessary horizontal speed (\(v_x\)) using the formula for horizontal distance: \(d = v_x \, t\). We have \(d = 3.5 \, \text{m}\) and \(t \approx 1.35 \, \text{s}\), so: \[ v_x = \frac{3.5 \, \text{m}}{1.35 \, \text{s}}. \]

Solution for Launch Speed

Finish the calculation to find the necessary horizontal speed: \[ v_x \approx 2.59 \, \text{m/s}. \] Thus, the launch speed needed to hit the bull's-eye is about \(2.59 \, \text{m/s}\).

Key concepts

Horizontal Launch

A horizontal launch occurs when an object is projected along a horizontal path from a certain height, without any initial angle. In our exercise, this is the situation when the pumpkin is thrown off the tower with a launch speed that we need to find. Here, the only force acting on the pumpkin after launch is gravity, as there is no vertical initial velocity. Thus, the path of the pumpkin is influenced by its initial horizontal speed.

To understand horizontal launch, it's crucial to note:

• The horizontal and vertical motions are independent of each other, meaning gravity only affects the vertical motion.
• The horizontal velocity remains constant because no force acts in the horizontal direction (ignoring air resistance).
• The key factor is calculating the time it takes for the pumpkin to hit the ground, as it affects how far it will travel horizontally.

Understanding these points helps us predict the projectile's path and the necessary initial speed.

Gravity

Gravity is the force that pulls objects toward the center of the Earth, affecting their vertical motion. For any object in free fall, gravity causes it to accelerate downwards at approximately \( 9.81 \, \frac{m}{s^2} \).

In the case of our horizontally launched pumpkin, gravity is the reason it eventually hits the ground. It acts continually on the pumpkin, increasing its vertical velocity as it falls. Importantly, for a horizontal launch:

• Gravity does not influence the horizontal speed of the projectile directly.
• The vertical motion of the projectile can be analyzed separately using the equations of motion.
• Without gravity, the pumpkin would continue to move in a straight line horizontally.

Recognizing the role of gravity helps predict how long it will take for an object to fall from a given height.

Equations of Motion

The equations of motion provide essential tools for predicting projectile behavior. In the case of the pumpkin, we utilize them to determine both the time of fall and the correct launch speed.

1. **Vertical motion equation:** This is given by \( h = \frac{1}{2} g t^2 \), where \( h \) is the height, \( g \) the acceleration due to gravity, and \( t \) the time. This determines how long the pumpkin takes to fall.2. **Horizontal motion equation:** Horizontal distance \( d \) is calculated by \( d = v_x t \), with \( v_x \) as the horizontal velocity. This lets us find the necessary launch speed for hitting the target.

These mathematical formulas allow for breaking down motion into simpler components, making it easier to find variables like launch speed.

Time of Fall

Time of fall is critical in projectile motion as it determines both vertical and horizontal distances. Calculating this involves solving the equation for vertical motion: \( h = \frac{1}{2} g t^2 \).

For the pumpkin thrown from the tower:

• The tower’s height is 9.0 m and gravity is \( 9.81 \, \frac{m}{s^2} \).
• By solving \( 9.0 = \frac{1}{2} \times 9.81 \times t^2 \), you find \( t \approx 1.35 \, s \).

This time affects how far horizontally the pumpkin will travel before hitting the ground. Identifying the time of fall helps in calculating the correct horizontal launch speed needed to hit a designated point like the bull’s-eye.