Problem 228 - Calculus Volume 1

Question

A ball is thrown upward from a height of \(3 \mathrm{~m}\) at an initial speed of \(60 \mathrm{~m} / \mathrm{sec}\). Acceleration resulting from gravity is \(-9.8 \mathrm{~m} / \mathrm{sec}^{2}\). Neglecting air resistance, solve for the velocity \(v(t)\) and the height \(h(t)\) of the ball \(t\) seconds after it is thrown and before it returns to the ground.

Step-by-Step Solution

Answer

Velocity: \(v(t) = -9.8t + 60\); Height: \(h(t) = -4.9t^2 + 60t + 3\).

Step 1: Determine the Velocity Function v(t)

The velocity of the ball as a function of time can be determined by integrating the acceleration due to gravity. Given the constant acceleration: \( a(t) = -9.8 \, \text{m/s}^2 \), the velocity function is found by integrating this with respect to time:\[\int a(t) \, dt = \int -9.8 \, dt = -9.8t + C_1\]We know the initial velocity \( v(0) = 60 \, \text{m/s} \). Substitute to find \( C_1 \):\[-9.8 \times 0 + C_1 = 60 \, \Rightarrow C_1 = 60\]Thus, the velocity function is:\[v(t) = -9.8t + 60\]

Step 2: Determine the Height Function h(t)

The height of the ball as a function of time can be determined by integrating the velocity function. We have:\[\int v(t) \, dt = \int (-9.8t + 60) \, dt = -4.9t^2 + 60t + C_2\]Given the initial height \( h(0) = 3 \, \text{m} \), substitute to find \( C_2 \):\[-4.9 \times 0^2 + 60 \times 0 + C_2 = 3 \, \Rightarrow C_2 = 3\]Therefore, the height function is:\[h(t) = -4.9t^2 + 60t + 3\]

Step 3: Verification and Conclusion

Verify by checking initial conditions: - For the velocity, substitute \( t = 0\) into \(v(t) = -9.8t + 60 \) to get \( v(0) = 60 \, \text{m/s} \), which is correct.- For the height, substitute \( t = 0 \) into \( h(t) = -4.9t^2 + 60t + 3 \) to get \( h(0) = 3 \, \text{m} \), which is also correct.Therefore, the functions are verified as: - Velocity: \( v(t) = -9.8t + 60 \)- Height: \( h(t) = -4.9t^2 + 60t + 3 \)

Key Concepts

Velocity Function

Velocity is a fundamental concept in kinematics, describing the rate at which an object's position changes over time. In this exercise, we start with the known acceleration due to gravity, which is a constant \(-9.8 \, \text{m/s}^2\). A negative sign indicates that gravity accelerates objects downward.

To find the velocity function, we need to integrate the acceleration function. Integration is the process of finding a function whose derivative is given. Here, we integrate:

• \[\int -9.8 \, dt = -9.8t + C_1\]

Given that the initial velocity when the ball is thrown upward is \(60 \, \text{m/s}\), we use this initial condition to solve for \(C_1\): \(-9.8 \cdot 0 + C_1 = 60\), which gives \(C_1\) as 60. Thus, the velocity function becomes: \[v(t) = -9.8t + 60\]
This formula allows us to calculate the ball's velocity at any time \(t\) before it hits the ground again.

Height Function

The height function describes how the vertical position of an object changes over time in kinematics. The height of the ball depends on its initial height, its velocity, and the acceleration due to gravity that we derived.

To find the height function, we integrate the velocity function, \(v(t) = -9.8t + 60\):

• \[\int (-9.8t + 60) \, dt = -4.9t^2 + 60t + C_2\]

Here, \(C_2\) is the constant of integration, which we determine using the initial condition that the ball is at a height of \(3 \, \text{m}\) when \(t = 0\):

• \(-4.9 \cdot 0^2 + 60 \cdot 0 + C_2 = 3 \Rightarrow C_2 = 3\)

Therefore, the height function is: \[h(t) = -4.9t^2 + 60t + 3\]
This equation captures the position of the ball at any time \(t\) during its motion, up until it returns to the ground.

Integration

Integration is a key mathematical tool used in kinematics to determine physical quantities that depend on rates of change. It is essentially the reverse process of differentiation.

In this problem, integration allows us to move from acceleration, a rate of change of velocity, to the velocity function, and from there to the height function. Each integration step requires adding a constant, often determined by initial conditions:

• From acceleration \(a(t) = -9.8\), we integrate to find velocity \(v(t) = -9.8t + C_1\).
• Next, integration of velocity \(v(t)\) gives us the height function \(h(t)\).

A constant derived from known values like initial speed or position helps in solving for the integration constant. Integration transforms the problem from a dynamic rate (like velocity) to terms like distance or position, which are more understandable in practical scenarios.

Gravity

Gravity is a force that causes an acceleration of \(-9.8 \, \text{m/s}^2\) towards Earth's surface. This force affects every object with mass.

In kinematics exercises like this one, the gravitational acceleration serves as a foundation for calculating motion characteristics such as velocity and height. Since we neglect air resistance, gravity remains the only force acting on the ball.

• The negative sign in \(-9.8 \, \text{m/s}^2\) indicates that gravity pulls objects downward, opposing any upward motion.
• It means that any object, once thrown up, will slow down, stop at its peak, and then speed up in a downward direction.

Understanding gravity and its effects explains why the velocity decreases over time during ascent and why the height function forms a parabolic path.