Problem 51
Question
Use \(a(t)=-32\) feet per second per second as the acceleration due to gravity. A ball is thrown vertically upward from a height of 6 feet with an initial velocity of 60 feet per second. How high will the ball go?
Step-by-Step Solution
Verified Answer
The maximum height that the ball will reach is 56.25 feet.
1Step 1: Recognise the given parameters
The problem provides these parameters: initial velocity \(v_0 = 60\) feet per second, initial height \(h_0 = 6\) feet, and acceleration \(a = -32\) feet per second per second (indicating downwards acceleration). The unknown parameter is the maximum height \(h_{max}\) the ball will attain.
2Step 2: Apply the principle of motion
The maximum height is attained when the velocity of the ball becomes zero, i.e., \(v = 0\). This is due to the fact that the ball changes its direction of motion at the maximum height. We can use the equation \(v^2 = v_0^2 + 2a(h - h_0)\), which is derived from the basic principles of motion under constant acceleration. At \(h_{max}\), the velocity \(v\) is zero. So, substituting the given values, we get \(0 = 60^2 + 2*-32*(h_{max} - 6)\).
3Step 3: Solve the equation
By solving the equation, we find \(h_{max} = 60^2/(2*-32) + 6 = 56.25\) feet.
Key Concepts
KinematicsAcceleration Due to GravityMaximum Height Calculation
Kinematics
Kinematics is the branch of physics that describes the motion of objects without considering the forces that cause the motion. It involves the relationships between displacement, velocity, and acceleration, mainly through time.
The core equations of kinematics help us answer questions about objects moving in linear paths, such as straight up or down, like our ball.
These kinematic equations are powerful tools for describing the motion of objects under uniform acceleration, such as the downward pull of gravity.
The core equations of kinematics help us answer questions about objects moving in linear paths, such as straight up or down, like our ball.
- Displacement is the change in position and is often denoted as \(s\).
- Velocity is the rate of change of displacement with time.
- Acceleration is the rate of change of velocity with time.
These kinematic equations are powerful tools for describing the motion of objects under uniform acceleration, such as the downward pull of gravity.
Acceleration Due to Gravity
Acceleration due to gravity is a constant that explains why objects fall towards Earth. It's denoted by the symbol \(g\) and typically has a value of \(-9.8\) meters per second squared. However, in our exercise using feet as the unit, it is \(-32\) feet per second squared.
The negative sign indicates the direction of acceleration: towards the Earth's surface. When an object is thrown upwards, like our ball, it experiences this downward pull, which gradually slows it down until it halts temporarily at its highest point before descending.
The negative sign indicates the direction of acceleration: towards the Earth's surface. When an object is thrown upwards, like our ball, it experiences this downward pull, which gradually slows it down until it halts temporarily at its highest point before descending.
- This constant acceleration affects all objects, regardless of their mass.
- It's crucial for calculations involving free-falling objects or those projected upwards.
Maximum Height Calculation
Calculating the maximum height an object reaches is a common task in physics problems involving vertical motion under gravity. This height is reached when the object's velocity becomes zero as it can no longer ascend against the pull of gravity.
To find the maximum height, we use the kinematic equation: \[v^2 = v_0^2 + 2a(h - h_0)\]Here, \(v\) is the final velocity (zero at maximum height), \(v_0\) is the initial velocity, \(a\) is the acceleration, and \(h - h_0\) represents the change in height. By rearranging, we solve for the unknown height.
Let's apply this to our exercise:
To find the maximum height, we use the kinematic equation: \[v^2 = v_0^2 + 2a(h - h_0)\]Here, \(v\) is the final velocity (zero at maximum height), \(v_0\) is the initial velocity, \(a\) is the acceleration, and \(h - h_0\) represents the change in height. By rearranging, we solve for the unknown height.
Let's apply this to our exercise:
- Initial velocity \(v_0 = 60\) feet/second
- Acceleration \(a = -32\) feet/second/second
- Initial height \(h_0 = 6\) feet
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Problem 51
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