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Magazine Chapter 2: Problem 60
A rock is thrown vertically upward from ground level at time \(t=0 .\) At \(t=1.5 \mathrm{~s}\) it passes the top of a tall tower, and \(1.0 \mathrm{~s}\) later it reaches its maximum height. What is the height of the tower?
Short Answer
Expert verified
The height of the tower is 25.725 meters.
Step by step solution
01
Understand the problem
We need to find the height of the tower that the rock passes at 1.5 seconds. The rock is thrown upward and continues to rise for another 1 second to reach its maximum height. We assume the rock is thrown from the base of the tower.
02
Identify the equations of motion
The rock's motion can be described by the equations of uniformly accelerated motion. The two relevant equations are: 1. The velocity equation: \( v = u + at \)2. The position equation: \( s = ut + \frac{1}{2}at^2 \)where \( u \) is the initial velocity, \( a \) is the acceleration (gravity, which is \( -9.8 \text{ m/s}^2 \) when upward), \( t \) is the time, and \( s \) is the displacement.
03
Determine initial velocity using maximum height condition
At maximum height (\( t = 2.5 \text{ s} \)), the velocity \( v = 0 \). Using \( v = u + at \), where \( a = -9.8 \text{ m/s}^2 \) and \( t = 2.5 \text{ s} \), we get:\( 0 = u - 9.8 \times 2.5 \)Solving for \( u \), we find the initial velocity \( u = 24.5 \text{ m/s} \).
04
Calculate displacement at 1.5 seconds
Substitute \( u = 24.5 \text{ m/s} \), \( a = -9.8 \text{ m/s}^2 \), and \( t = 1.5 \text{ s} \) into the position equation \( s = ut + \frac{1}{2}at^2 \):\[ s = 24.5 \times 1.5 + \frac{1}{2}(-9.8) \times (1.5)^2 \]Calculate to get:\[ s = 36.75 - 11.025 = 25.725 \text{ m} \]
05
Conclude the height of the tower
The height of the tower, which the rock passes at \( t = 1.5 \text{ s} \), is \( 25.725 \text{ m} \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Equations of Motion
In physics, the equations of motion are crucial for analyzing the movement of objects. These equations describe how an object's position and velocity change with time. For this exercise, we use two primary equations:
These equations are applied universally to problems where uniform acceleration is present, like a rock thrown in the air, as in our given problem.
Knowing how to manipulate these equations helps solve complex motion-related issues.
- The velocity equation, given by: \( v = u + at \)
- The position equation, or displacement formula: \( s = ut + \frac{1}{2}at^2 \)
These equations are applied universally to problems where uniform acceleration is present, like a rock thrown in the air, as in our given problem.
Knowing how to manipulate these equations helps solve complex motion-related issues.
Uniformly Accelerated Motion
When dealing with motion under constant acceleration, we refer to it as uniformly accelerated motion. This concept is depicted well when any object, like a rock in our scenario, moves under the influence of gravity.
Gravity is an ever-present force on Earth, constantly exerting an acceleration of \( -9.8 \text{ m/s}^2 \) on objects. This negative sign indicates that gravity pulls objects downwards.
Gravity is an ever-present force on Earth, constantly exerting an acceleration of \( -9.8 \text{ m/s}^2 \) on objects. This negative sign indicates that gravity pulls objects downwards.
- Uniform acceleration means that the acceleration value does not change throughout the motion.
- All the equations of motion can be applied to uniformly accelerated systems.
Initial Velocity Calculation
The initial velocity is the speed at which an object begins its motion. To find this, especially in situations involving projectiles, we can use the equations of motion.
In our exercise, the rock reaches its maximum height at 2.5 seconds, implying a temporary velocity of zero at that point.
By setting the final velocity \( v = 0 \) in the equation \( v = u + at \), and knowing the acceleration is \( -9.8 \text{ m/s}^2 \) and time taken is 2.5 seconds, we solved for the initial velocity:
In our exercise, the rock reaches its maximum height at 2.5 seconds, implying a temporary velocity of zero at that point.
By setting the final velocity \( v = 0 \) in the equation \( v = u + at \), and knowing the acceleration is \( -9.8 \text{ m/s}^2 \) and time taken is 2.5 seconds, we solved for the initial velocity:
- \( 0 = u - 9.8 \times 2.5 \)
- Resulting in \( u = 24.5 \text{ m/s} \)
Displacement Calculation
Displacement refers to an object's overall change in position. We calculate it using the position equation: \( s = ut + \frac{1}{2}at^2 \). When the rock is thrown, it covers a certain vertical distance until it passes the tower at 1.5 seconds.
Displacement can tell us how far an object travels, which is essential for understanding its journey. In context, it helps determine the height of the tower.
- Using the initial velocity \( u = 24.5 \text{ m/s} \),
- the acceleration \( a = -9.8 \text{ m/s}^2 \),
- and time \( t = 1.5 \text{ s} \),
Displacement can tell us how far an object travels, which is essential for understanding its journey. In context, it helps determine the height of the tower.
