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A projectile is launched from and returns to ground level, as the figure shows. There is no air resistance. The horizontal range of the projectile is measured to be \(R=175 \mathrm{m},\) and the horizontal component of the launch velocity is \(v_{0 x}=+25 \mathrm{m} / \mathrm{s} .\) Concepts: (i) What is the final value of the horizontal component \(v_{x}\) of the projectile's velocity? (ii) Can the time be determined for the horizontal part of the motion? (iii) Is the time for the horizontal part of the motion the same as that for the vertical part? (iv) For the vertical part of the motion, what is the displacement of the projectile? Calculations: Find the vertical component \(v_{0 y}\) of the projectile.

Step by step solution

01

Identify the horizontal component of velocity

The horizontal component of the projectile's velocity remains constant throughout the motion due to the absence of air resistance. Therefore, the final horizontal velocity \( v_x \) is the same as the initial velocity: \( v_x = v_{0x} = +25 \, \text{m/s} \).

02

Determine the time of flight using horizontal motion

For the horizontal motion, we can determine the time of flight \( t \) using the horizontal range \( R \) and horizontal velocity \( v_{0x} \). The formula used is: \[ t = \frac{R}{v_{0x}} = \frac{175 \, \text{m}}{25 \, \text{m/s}} = 7 \, \text{s} \].

03

Confirm time of flight consistency

The time for the horizontal part of the motion is indeed the same as for the vertical part due to the projectile returning to the ground level at the end of its flight. Thus, vertical and horizontal motions share the same time of flight.

04

Determine the vertical displacement for vertical motion

For vertical motion, the displacement \( y \) is zero because the projectile returns to the same vertical position it was launched from.

05

Calculate the vertical component of initial velocity

Using the time of flight and the gravitational acceleration \( g = 9.8 \, \text{m/s}^2 \), find the vertical component using the equation: \[ v_{0y} = \frac{g \, t}{2} = \frac{9.8 \, \text{m/s}^2 \times 7 \, \text{s}}{2} = 34.3 \, \text{m/s} \].

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Horizontal Range

The horizontal range of a projectile, often just called "range," is how far it travels along the horizontal axis. This range is determined by both the horizontal speed and the total time the projectile spends in motion. In this specific exercise, the range is given as 175 meters. It can be calculated using the formula:

• \( R = v_{0x} \times t \)

where \( v_{0x} \) is the horizontal component of the initial velocity, and \( t \) is the time of flight.
If you know the horizontal range and horizontal velocity, you can work backwards to find the time of flight for any projectile motion, assuming no air resistance is affecting the motion.

Vertical Displacement

Vertical displacement in projectile motion refers to the change in vertical position from the starting point to the ending point. In our case, because the projectile returns to the ground, the starting and ending vertical positions are the same. This means that the vertical displacement is zero.
This might sound surprising because the projectile does indeed rise and fall. However, vertical displacement specifically measures the net change in position—not the actual distance traveled up and down. By focusing solely on starting and ending points, we see there was no net change in vertical position as it returned to ground level.

Time of Flight

Time of flight is the total time a projectile spends in air from launch until landing. The time of flight is the same for both the horizontal and vertical components of motion. In this case, the time of flight was determined using the horizontal range and velocity:

• \( t = \frac{R}{v_{0x}} \)

With a range of 175 meters and a horizontal velocity of 25 m/s, the time calculated is 7 seconds.
For projectile motion problems, especially when returning to the original height, this shared time value is crucial in calculating other aspects like maximum height or initial vertical velocity.

Initial Velocity Components

The initial velocity of a projectile can be broken down into two components: horizontal and vertical. For the horizontal component, the problem statement provides \( v_{0x} = 25 \, \text{m/s} \). This velocity remains constant in projectile motion without air resistance.
The vertical component \( v_{0y} \) is less straightforward since it changes due to gravity. We calculated \( v_{0y} \) using the total time of flight and gravitational acceleration \( g = 9.8 \, \text{m/s}^2 \). Here, the formula is:

• \( v_{0y} = \frac{g \times t}{2} \)

where \( t = 7 \text{s} \). Using this, we found \( v_{0y} = 34.3 \, \text{m/s} \). These two components together define the projectile's path and behavior throughout its flight.