Question

Consider the following descriptions of the vertical motion of an object subject only to the acceleration due to gravity. Begin with the acceleration equation \(a(t)=v^{\prime}(t)=g,\) where \(g=-9.8 \mathrm{m} / \mathrm{s}^{2}\) a. Find the velocity of the object for all relevant times. b. Find the position of the object for all relevant times. c. Find the time when the object reaches its highest point. What is the height? d. Find the time when the object strikes the ground. A stone is thrown vertically upward with a velocity of \(30 \mathrm{m} / \mathrm{s}\) from the edge of a cliff \(200 \mathrm{m}\) above a river.

Short answer

Answer: The object reaches its highest point at approximately t = 3.06 seconds with a height of 245.4 meters.

Step-by-step solution

a. Find the velocity of the object for all relevant times.

First, let's integrate the acceleration equation with respect to time to find the velocity equation. Since \(a(t) = v'(t) = g\): $$v(t) = \int a(t) \, dt = \int (-9.8) \, dt = -9.8t + C$$ Since the initial velocity is \(30\,\mathrm{m}/\mathrm{s}\), we have \(v(0) = 30\), and we can solve for the integration constant \(C\). Thus, the velocity equation is: $$v(t) = -9.8t + 30$$

b. Find the position of the object for all relevant times.

Next, let's integrate the velocity equation with respect to time to find the position equation. Since \(v(t) = s'(t)\): $$s(t) = \int v(t) \, dt = \int (-9.8t + 30) \, dt = -4.9t^2 + 30t + D$$ Since the initial height is \(200\,\mathrm{m}\), we have \(s(0) = 200\), and we can solve for the integration constant \(D\). Thus, the position equation is: $$s(t) = -4.9t^2 + 30t + 200$$

c. Find the time when the object reaches its highest point. What is the height?

To find the highest point, we need to find when the velocity is equal to zero: $$0 = -9.8t + 30 \Rightarrow t = \frac{30}{9.8} \approx 3.06\,\mathrm{s}$$ Now, we substitute this time into the position equation to find the height at the highest point: $$s(3.06) = -4.9(3.06)^2 + 30(3.06) + 200 \approx 245.4\,\mathrm{m}$$ So, the highest point is reached at \(t \approx 3.06 \, \mathrm{s}\) with a height of \(245.4\,\mathrm{m}\).

d. Find the time when the object strikes the ground.

To find the time when the object hits the ground, we need to find when the position equation is equal to zero: $$0 = -4.9t^2 + 30t + 200$$ This is a quadratic equation, and we can use the quadratic formula to find the roots: $$t = \frac{-30 \pm \sqrt{(-30)^2 - 4(-4.9)(200)}}{2(-4.9)}$$ We'll get two values for \(t\), but since it was thrown upward, we're only interested in the positive value. We can solve for \(t\): $$t \approx 6.35\,\mathrm{s}$$ So, the object strikes the ground at \(t \approx 6.35\,\mathrm{s}\).

Key concepts

Integration in Physics

Integration is a fundamental concept in physics, especially when dealing with motion under the influence of constant forces. It allows us to predict positions, velocities, and acceleration over time based on initial conditions. In the context of vertical motion calculus problems, integration is used to determine how an object's velocity and position change over time when it is subject only to the force of gravity.

In our exercise example, the process begins by integrating the constant acceleration due to gravity, represented by the symbol 'g', to get the velocity function \( v(t) \). Once we have the velocity as a function of time, a second integration gives us the position function \( s(t) \), which tells us the location of the object at any given time. Each integration requires determining a constant of integration by using the given initial conditions, such as initial velocity or initial position.

For instance, consider the initial velocity of \(30 \text{m/s}\) in the example. This is the condition used to solve for an integration constant when finding the velocity equation from the acceleration equation. This stepwise integration from acceleration to velocity, and then to position, adheres to the principles of calculus and is fundamental in the study of kinematics in physics.

Acceleration Due to Gravity

The acceleration due to gravity, commonly denoted as \(g\), is a constant force that acts on objects near the Earth's surface, pulling them towards the center of the Earth. This acceleration is approximately \(9.8 \text{m/s}^2\) and is always directed downwards. It is an essential factor in the study of projectile motion or any movement involving free fall.

In our example, gravity is the sole force acting on the object, and it gives the object a constant acceleration of \(g = -9.8 \text{m/s}^2\). The negative sign indicates that gravity is acting in the opposite direction to the positive displacement (upwards direction). When an object is thrown vertically upwards, gravity will slow it down until it reaches its highest point, where the velocity momentarily becomes zero before the object descends back to the ground, accelerating under gravity.

Understanding how gravity influences motion is key for predicting the behavior of objects in free fall—calculating the time it takes to reach the highest point, the maximum height attained, and the time required to hit the ground are all applications of this concept. The acceleration due to gravity impacts every term in our position equation \( s(t) \) and is foundational to the motion equations generated by integration.

Quadratic Equations

Quadratic equations play a crucial role in solving problems involving parabolic motion, which is characteristic of objects in free fall under the influence of gravity. In a quadratic equation, the highest exponent for the variable is two, and it has the general form \( ax^2 + bx + c = 0 \), where \(a\), \(b\), and \(c\) are constants, and \(x\) is the unknown variable we wish to solve for. These equations often appear in physics when finding the time at which certain events occur, such as reaching the highest point or striking the ground.

The position equation \( s(t) = -4.9t^2 + 30t + 200 \) from our exercise is a perfect example of a quadratic equation in which \( t \) is the variable representing time. To find the time when the object hits the ground, we set \( s(t) \) equal to zero and solve the resulting quadratic equation for \( t \). The quadratic formula, \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) is a reliable method for finding the time values that solve this equation. In our case, it gives us two values for \( t \), but we choose the positive value as it corresponds to the time after the stone is thrown upwards. Understanding how to interpret the solutions to quadratic equations is essential for analyzing the motion of objects under gravity.