Question

At the end of the spring term, a high school physics class celebrates by shooting a bundle of exam papers into the town landfill with a homemade catapult. They aim for a point that is \(30.0 \mathrm{~m}\) away and at the same height from which the catapult releases the bundle. The initial horizontal velocity component is \(3.90 \mathrm{~m} / \mathrm{s}\). What is the initial velocity component in the vertical direction? What is the launch angle?

Short answer

Answer: The initial vertical velocity is 37.7 m/s, and the launch angle is approximately 84.2°.

Step-by-step solution

Calculate time of flight

Using the first equation: \(x = x_0 + v_{0x}t_{\text{flight}}\) Since the initial position \(x_0 = 0\), we can write: \(t_{\text{flight}} = \frac{x}{v_{0x}}\) Plugging in the given values, we get: \(t_{\text{flight}} = \frac{30.0 \mathrm{~m}}{3.90 \mathrm{~m} / \mathrm{s}} \approx 7.69 \mathrm{~s}\)

Calculate initial vertical velocity

Using the second equation: \(0 = v_{0y}t_{\text{flight}} - \frac{1}{2}gt^2_{\text{flight}}\) Solving for \(v_{0y}\), we get: \(v_{0y} = \frac{1}{2}gt_{\text{flight}}\) Using \(g = 9.81 \mathrm{~m} / \mathrm{s}^2\) and \(t_{\text{flight}} \approx 7.69 \mathrm{~s}\), we calculate \(v_{0y}\): \(v_{0y} = \frac{1}{2}(9.81 \mathrm{~m} / \mathrm{s}^2)(7.69 \mathrm{~s}) \approx 37.7 \mathrm{~m} / \mathrm{s}\)

Calculate launch angle

Using the third equations, we get \(\tan{\theta} = \frac{v_{0y}}{v_{0x}}\) Plugging in the known values: \(\tan{\theta} = \frac{37.7 \mathrm{~m} / \mathrm{s}}{3.90 \mathrm{~m} / \mathrm{s}} \approx 9.67\) Now, compute the launch angle: \(\theta = \arctan{(9.67)} \approx 84.2^\circ\) So, the initial vertical velocity component is \(37.7\mathrm{~m} /\mathrm{s}\), and the launch angle is approximately \(84.2^\circ\).

Key concepts

Horizontal Velocity in Projectile Motion

Understanding horizontal velocity is crucial when analyzing projectile motion. It refers to how fast an object moves along the horizontal axis from the point of release. In projectile motion, horizontal velocity remains constant if air resistance is ignored.
Think about the initial problem where the exam papers are hurled towards a landfill. With an initial horizontal velocity of \(3.90 \text{ m/s}\), this speed doesn't change during the flight. That's because only gravity affects vertical motion, not horizontal.
This consistent horizontal speed helps us figure out how long the journey takes by using the formula:

• \(t_{\text{flight}} = \frac{x}{v_{0x}}\)

By plugging in \(x = 30.0 \text{ m}\) and \(v_{0x} = 3.90 \text{ m/s}\), we find the time of flight is about \(7.69 \text{ s}\). This time reflects how long the projectile remains airborne due to its horizontal motion.

Vertical Velocity Component

Vertical velocity is just as vital as horizontal velocity when it comes to projectile motion. It represents the speed at which an object moves up or down. Unlike horizontal velocity, vertical velocity is influenced by gravity.
In the exercise, the vertical component of velocity is required to calculate how the object rises before descending back to the same height. At the topmost point, vertical velocity is zero momentarily before gravity pulls the object back down.
We use the formula:

• \(v_{0y} = \frac{1}{2}gt_{\text{flight}}\)

With the acceleration due to gravity \(g = 9.81 \text{ m/s}^2\) and \(t_{\text{flight}} \approx 7.69 \text{ s}\), the calculation yields an initial vertical velocity \(v_{0y} \approx 37.7 \text{ m/s}\). This value tells us how fast the bundle was initially moving upward.

Launch Angle and Its Significance

The launch angle is a key factor in projectile motion. It is the angle the initial velocity vector makes with the horizontal.
This angle critically influences the range and maximum height of the projectile. In the exercise, the bundle of papers is released at a launch angle of approximately \(84.2^\circ\), indicating a very steep trajectory.
You can determine the launch angle using:

• \(\tan{\theta} = \frac{v_{0y}}{v_{0x}}\)

Converting the ratio of the initial vertical and horizontal velocities, we find \(\theta = \arctan(9.67)\), leading to a launch angle of about \(84.2^\circ\).
This angle suggests the projectile will ascend almost straight up, before coming down, traveling mostly vertically. Understanding this concept helps us predict the behavior and impact point of the projectile.