Problem 46 - Calculus Early Transcendentals: Pearson New International Edition

Question

A ball is thrown vertically upward with velocity \(v_{0}\). Find the maximum height \(H\) of the ball as a function of \(v_{0}\). Then find the velocity \(v_{0}\) required to achieve a height of \(H .\) Hint: The height of the ball after \(t\) seconds is \(h=-16 t^{2}+v_{0} t .\) The vertex of the parabola \(y=-a x^{2}+b x\) is at \(\left(b /(2 a), b^{2} /(4 a)\right)\).

Step-by-Step Solution

Answer

The maximum height is \(\frac{v_0^2}{64}\) and the required initial velocity is \(8\sqrt{H}\).

Step 1: Understand the Problem

We need to find the maximum height of a ball thrown vertically upwards with an initial velocity \(v_0\). The ball's height as a function of time \(t\) is given by \(h = -16t^2 + v_0 t\). When dealing with a quadratic function of the form \(y = -ax^2 + bx\), the vertex of the parabola provides the maximum value along the trajectory, as the leading coefficient is negative.

Step 2: Identify the Vertex Formula

Recall that the vertex of the quadratic function \(y = -ax^2 + bx\) is given by the point \(\left( \frac{b}{2a}, \frac{b^2}{4a} \right)\). This form is useful to determine the time \(t\) at which the ball reaches its maximum height, as well as the maximum height itself.

Step 3: Apply the Vertex Formula

For the equation \(h = -16t^2 + v_0 t\), the "a" coefficient is \(-16\) and the "b" coefficient is \(v_0\). Thus, the time \(t\) at which the maximum height is reached is \(t = \frac{v_0}{2(-16)} = \frac{v_0}{-32}\).

Step 4: Calculate Maximum Height with \(t\)

Substitute \(t = \frac{v_0}{-32}\) back into the equation for height: \(h = -16\left( \frac{v_0}{-32} \right)^2 + v_0 \left( \frac{v_0}{-32} \right)\). Simplifying this, \(h = \frac{v_0^2}{64} + \frac{-v_0^2}{32} = \frac{v_0^2}{64}\). Therefore, the maximum height \(H\) is \(\frac{v_0^2}{64}\).

Step 5: Solve for Initial Velocity \(v_0\) With Given Maximum Height

Given the maximum height \(H\), we have \(H = \frac{v_0^2}{64}\). Solve for \(v_0\) by multiplying both sides by 64: \(64H = v_0^2\), then take the square root: \(v_0 = \sqrt{64H} = 8\sqrt{H}\).

Key Concepts

Quadratic Function

A quadratic function is a type of polynomial function specifically having degree 2. Its general form is \(y = ax^2 + bx + c\). In our exercise, the function that models the height of the ball is \(h = -16t^2 + v_0 t\), showing the variables of time \(t\) and initial velocity \(v_0\). This equation lacks the constant term \(c\), indicating the ball is launched from ground level.

Quadratic functions are graphically represented in the form of a parabola, which is a U-shaped curve. Depending on the sign of coefficient \(a\), the parabola can open upwards or downwards. With our model's \(a = -16\) (since it's negative), the parabola opens downwards. This implies the height of the ball increases to a certain maximum point (vertex) before descending back to the ground.

Understanding quadratic functions is essential as they model various real-world phenomena, particularly those involving projectile motion, such as throwing a ball.

Vertex Formula

The vertex formula is crucial in finding the maximum or minimum values of a quadratic function. For the standard quadratic \(y = ax^2 + bx + c\), the vertex point \((x, y)\) is given by:

• \(x = \frac{-b}{2a}\)
• \(y = \frac{b^2 - 4ac}{4a}\)

In the context of our exercise, the vertex signifies the highest point the ball reaches during its flight. We can easily substitute \(-16\) for \(a\) and \(v_0\) for \(b\) to calculate the time \(t\) for maximum height:

• \(t = \frac{v_0}{-32}\)

By plugging this into the height equation, we determined the maximum height \(H = \frac{v_0^2}{64}\).

Grasping the vertex formula helps inform the optimization of energy or design in different types of projectile motion.

Initial Velocity

Initial velocity, often denoted as \(v_0\), is the speed at which an object begins its motion. For the ball in this exercise, it's the velocity at which it is thrown upward. This initial push dictates how high the ball can rise before gravity takes over and pulls it back down.

Consider this: higher initial velocity means a higher maximum height. Our task involves not just finding maximum height \(H\) but also reversing that work to determine the initial velocity required to achieve a certain height. Using the derived formula, you can establish the initial velocity with:

• \(v_0 = 8\sqrt{H}\)

The initial velocity is critical in real-world applications such as sports, rocketry, or any field where predictability of motion is needed.

Parabolic Motion

Parabolic motion describes the path of motion under the influence of uniform gravity, generally forming a parabola. When you throw an object into the air, like a ball, gravity accelerates it downwards, creating this distinct path.

In our exercise's context, the ball's motion is vertical, a simplified form of projectile motion. The parabolic motion equation \(h = -16t^2 + v_0 t\) outlines how height varies with time. The initial ascent slows due to gravity until reaching a brief period of zero velocity at the top, then descends back.

This idea of parabolic motion extends into multiple realms, like physics, engineering, and even animation, wherever gravity plays a key role in determining object paths.