Problem 23 - Fundamentals of Physics

Question

A projectile is fired horizontally from a gun that is \(45.0 \mathrm{~m}\) above flat ground, emerging from the gun with a speed of \(250 \mathrm{~m} / \mathrm{s}\). (a) How long does the projectile remain in the air? (b) At what horizontal distance from the firing point does it strike the ground? (c) What is the magnitude of the vertical component of its velocity as it strikes the ground?

Step-by-Step Solution

Answer

The projectile is in the air for about 3.03 s, travels 757.5 m horizontally, and hits the ground with a vertical speed of about 29.7 m/s.

Step 1: Analyze the problem

We need to find how long the projectile remains in the air (Part a), how far it travels horizontally (Part b), and the vertical velocity component at impact (Part c). Given the initial height is \(45.0\, \text{m}\), and the horizontal speed is \(250\, \text{m/s}\). Since it is fired horizontally, the initial vertical velocity is \(0\).

Step 2: Calculate the time of flight (a)

The time the projectile stays in the air depends on the time it takes to fall \(45.0\, \text{m}\) vertically. Using the equation of motion for vertical displacement, \( h = \frac{1}{2}gt^2 \), where \(g\) is the acceleration due to gravity \( (9.8 \text{m/s}^2) \). Solve for \(t\): \[ 45.0 = \frac{1}{2} \times 9.8 \times t^2 \] \[ t^2 = \frac{90.0}{9.8} \approx 9.18 \] \[ t \approx \sqrt{9.18} \approx 3.03\, \text{s} \].

Step 3: Calculate the horizontal distance (b)

Since the horizontal velocity \(v_x\) is constant at \(250\, \text{m/s}\), and we found the time of flight is \(3.03\, \text{s}\), we can calculate the horizontal distance \(d_x\) using \(d_x = v_x \cdot t\). \[ d_x = 250 \times 3.03 = 757.5\, \text{m} \].

Step 4: Calculate the vertical velocity at impact (c)

Using the formula for final velocity in freefall \(v_y = g \cdot t\), where \(g = 9.8\, \text{m/s}^2\) and \(t = 3.03\, \text{s}\): \[ v_y = 9.8 \times 3.03 \approx 29.7\, \text{m/s} \].

Key Concepts

Horizontal Velocity

In projectile motion, the term "horizontal velocity" refers to the constant speed at which a projectile travels horizontally. It's important to note that this velocity does not change throughout the motion, assuming air resistance is negligible. For example, if a projectile is fired from a gun at a speed of \(250\, \text{m/s}\), this value remains unchanged until the projectile hits the ground.
The horizontal component of motion does not affect how long the projectile is in the air or the vertical descent due to gravity. Because there are no horizontal forces acting on the projectile, the horizontal velocity remains consistent:

• Initial horizontal velocity \(v_x = 250\, \text{m/s}\)
• This value remains the same until impact

This concept helps simplify calculations because one can separate vertical and horizontal components of motion when solving projectile problems.

Time of Flight

"Time of flight" is the duration in which a projectile remains in the air from the moment it is launched until it strikes the ground. In a projectile problem, this time is usually determined by analyzing the vertical motion, especially when the projectile is launched horizontally.
Here's how you calculate the time of flight using the formula for vertical displacement:

• Start with the equation: \( h = \frac{1}{2}gt^2 \)
• Solve for time \( t \) as it falls from a height of \(45\, \text{m}\)

By plugging in our values \(45 = \frac{1}{2} \times 9.8 \times t^2\), you find \( t \approx 3.03\, \text{s} \).
This tells us that the projectile remains in the air for approximately \(3.03\, \text{seconds}\), irrespective of its horizontal velocity.

Vertical Velocity

Vertical velocity in projectile motion refers to the speed at which the projectile travels upward or downward. When a projectile is launched horizontally, its initial vertical velocity is zero because it doesn't initially move up or down. However, as time progresses, the vertical velocity changes due to the pull of gravity. This is an example of constant acceleration motion.
To find the vertical velocity at impact, use the formula:

• \(v_y = gt\), where \(g = 9.8\, \text{m/s}^2\), and \(t\) is the time of flight.

Given \( t = 3.03\, \text{seconds}\),
Plug these into the equation: \( v_y = 9.8 \times 3.03 \approx 29.7\, \text{m/s}\).
This calculation indicates how fast the projectile is moving vertically downward upon impacting the ground.

Acceleration Due to Gravity

"Acceleration due to gravity" is a crucial concept in understanding projectile motion. On Earth, this value is a constant \(9.8\, \text{m/s}^2\), directed downward toward the center of the Earth. This acceleration affects only the vertical component of a projectile's motion, not the horizontal.
Here's why gravity is key in projectile calculations:

• Gravity causes the projectile to accelerate downward, increasing its vertical velocity over time.
• It dictates how long a projectile remains in the air during its time of flight.
• At impact, the vertical velocity is solely dependent on gravity and the time the projectile has been falling.

Thus, understanding gravity allows us to calculate both the time of flight and the final vertical velocity accurately, providing a comprehensive view of the object's motion throughout its trajectory.