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You can use any coordinate system you like to solve a projectile motion problem. To demonstrate the truth of this statement, consider a ball thrown off the top of a building with a velocity \(\vec{v}\) at an angle \(\theta\) with respect to the horizontal. Let the building be \(50.0 \mathrm{~m}\) tall, the initial horizontal velocity be \(9.00 \mathrm{~m} / \mathrm{s}\), and the initial vertical velocity be \(12.0 \mathrm{~m} / \mathrm{s}\). Choose your coordinates such that the positive \(y\) -axis is upward, the \(x\) -axis is to the right, and the origin is at the point where the ball is released. (a) With these choices, find the ball's maximum height above the ground and the time it takes to reach the maximum height. (b) Repeat your calculations choosing the origin at the base of the building.

Step by step solution

01

Part A: Calculate maximum height and time to reach maximum height from top of the building

1. Calculate the time it takes to reach the maximum height using the formula \(t_{max} = V_{0y} / g\), where \(V_{0y}\) is the initial vertical velocity and \(g\) is acceleration due to gravity. We have \(V_{0y} = 12.0 \, m/s\) and \(g = 9.8 \, m/s^2\). Upon substituting these values, we get \(t_{max}\). \n 2. Using this \(t_{max}\), the maximum height \(h_{max}\) from the launching point can be calculated using the formula \(h_{max} = V_{0y} \cdot t_{max} - 0.5 \cdot g \cdot t_{max}^2\).

02

Part B: Calculate maximum height and time to reach maximum height from the base of the building

1. To calculate the maximum height from the base of the building, we add the height of the building \(H\) to \(h_{max}\) we computed above, as \[H_{total} = H + h_{max}\] where \(H = 50.0 \, m\). \n 2. The time to reach maximum height remains the same even if the origin is shifted because it is independent of the starting point. Thus, the time remains the same as \(t_{max}\) calculated in Part A.

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Key Concepts

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

Coordinate System

In projectile motion problems, choosing the right coordinate system simplifies your calculations. Consider the ball in motion. You can use any coordinate system, as long as you keep it consistent. Typically in projectile motion:

• The x-axis is horizontal.
• The y-axis is vertical and points upward.

The origin can be at the starting point of the projectile, like the point where the ball is released. Alternatively, it could be at a different location, like the base of the building.
Remember, the choice of the coordinate system does not change the physics involved. It only influences the calculation steps by altering your equations' constants. So, be clear with your axes directions and origin point to avoid confusion.

Initial Velocity

Initial velocity is crucial in determining a projectile's path. It's composed of two components:

• Horizontal initial velocity, \( V_{0x} \). This remains constant as no horizontal force acts on the projectile (assuming air resistance is negligible).
• Vertical initial velocity, \( V_{0y} \). This changes over time due to gravity.

For the given exercise:The initial horizontal velocity is \( 9.00 \ m/s \) and the initial vertical velocity is \( 12.0 \ m/s \).
The initial velocity determines how fast and in which direction the projectile starts moving. You can calculate it as a vector \( \vec{v} = (V_{0x}, V_{0y}) \). These components help in finding subsequent motion-related parameters.

Maximum Height

The maximum height of a projectile is the highest point it reaches during its flight. To find it, calculate the time at which the vertical velocity becomes zero. This is given by:\[ t_{max} = \frac{V_{0y}}{g} \]where \( V_{0y} \) is the initial vertical velocity, and \( g \) is acceleration due to gravity.
Once you know \( t_{max} \), you can use it to find the maximum height above the starting point:\[ h_{max} = V_{0y} \cdot t_{max} - 0.5 \cdot g \cdot t_{max}^2 \]If the origin is elsewhere, such as at the base of the building, simply add the building’s height to this result.
Understanding how height changes with different origins and knowing these equations, simplifies solving projectile motion problems.

Acceleration Due to Gravity

Acceleration due to gravity is a constant force acting on all objects near the Earth’s surface. It's approximately \( 9.8 \ m/s^2 \) downward. In projectile motion, gravity affects only the vertical component of velocity.
This force:

• Decelerates the projectile as it moves upward.
• Accelerates it downward until it hits the ground.

The consistent pull due to gravity means the horizontal velocity stays unaffected. Only the vertical motion experiences acceleration, which explains why we calculate maximum height and time to reach it using vertical velocity. Understanding gravity’s role is essential, as it consistently influences the entire motion.