Problem 71 - Calculus and its applications

Question

A baseball is thrown directly upward with an initial velocity of \(75 \mathrm{ft} /\) sec from an initial height of \(30 \mathrm{ft}\). The velocity of the baseball \(t\) seconds after being released is given by $$v(t)=-32 t+75$$ where \(v\) is in feet per second. a) Find the function \(h\) that gives the height (in feet) of the baseball after \(t\) seconds. b) What are the height and the velocity of the baseball after 2 sec of flight? c) After how many seconds does the ball reach its highest point? (Hint: The ball "stops" for a moment before starting its downward fall.) d) How high is the ball at its highest point? e) After how many seconds will the ball hit the ground? f) What is the ball's velocity at the moment it hits the ground?

Step-by-Step Solution

Answer

a) \(h(t) = -16t^2 + 75t + 30\). b) Height is 116 ft, velocity is 11 ft/s at 2 sec. c) Highest point at \(\frac{75}{32}\) sec. d) Use \(h(t)\) to find height. e) Solve for \(t\) when \(h(t) = 0\). f) Substitute this \(t\) into \(v(t)\).

Step 1: Understand Initial Conditions

The baseball is thrown from a 30 ft height with a velocity of 75 ft/s. The velocity function is given by \(v(t) = -32t + 75\), where \(-32\) ft/s² is the acceleration due to gravity.

Step 2: Derive Height Function h(t)

The velocity is the derivative of the height function \(h(t)\). Thus, integrate \(v(t) = -32t + 75\) to get \(h(t)\). This gives \(h(t) = \int (-32t + 75) \, dt = -16t^2 + 75t + C\). The initial height \(h(0) = 30\) allows us to solve for \(C\): \(C = 30\). Thus, \(h(t) = -16t^2 + 75t + 30\).

Step 3: Calculate Height and Velocity at 2 Seconds

Use \(h(t) = -16t^2 + 75t + 30\) to find the height at \(t = 2\): \(h(2) = -16\cdot2^2 + 75\cdot2 + 30 = -64 + 150 + 30 = 116\) ft. Calculate velocity with \(v(t) = -32t + 75\) at \(t = 2\): \(v(2) = -32\cdot2 + 75 = -64 + 75 = 11\) ft/s.

Step 4: Find When Ball Reaches Highest Point

The ball reaches its highest point when \(v(t) = 0\). Solve \(-32t + 75 = 0\) to get \(-32t = -75\). Thus, \(t = \frac{75}{32}\) seconds.

Step 5: Calculate Highest Point

Use \(h(t) = -16t^2 + 75t + 30\) to find \(h\left(\frac{75}{32}\right)\): \(h\left(\frac{75}{32}\right) = -16\left(\frac{75}{32}\right)^2 + 75\left(\frac{75}{32}\right) + 30\). Simplify this to find the maximum height.

Step 6: Determine When Ball Hits Ground

The ball hits the ground when \(h(t) = 0\). Solve \(-16t^2 + 75t + 30 = 0\) using the quadratic formula: \(t = \frac{-75 \pm \sqrt{75^2 - 4\cdot(-16)\cdot30}}{2\cdot(-16)}\). Discard the negative solution to find \(t\).

Step 7: Compute Velocity at Impact

Substitute the positive \(t\) from Step 6 into \(v(t) = -32t + 75\) to find the ball's velocity just before it hits the ground.

Key Concepts

Height Function

In projectile motion, the height function describes how high an object is above the ground over time. For a baseball thrown upwards, the height function, denoted as \( h(t) \), informs us of the height of the baseball at any given second.
To derive \( h(t) \) from the velocity function \( v(t) \), we integrate because the velocity function gives the rate of change of height. This integration results in

• \( h(t) = -16t^2 + 75t + C \)

Here, \(-16t^2\) accounts for the acceleration due to gravity (\(-32 \text{ ft/s}^2\)), and \(75t\) represents the initial upwards velocity. The constant \( C \) is determined using the initial height of the baseball, which is 30 feet. Therefore, \( C = 30 \). Finally, the height function is

• \( h(t) = -16t^2 + 75t + 30 \)

This quadratic equation now describes the height over time, taking into consideration the initial conditions.

Velocity Function

The velocity function gives us critical information about the speed and direction of the baseball as time progresses. For a projectile like a baseball, the velocity function is a linear equation representing how fast the ball moves upwards or downwards at any given second.
In this exercise, the velocity function is given by

• \( v(t) = -32t + 75 \)

Here's what each component means:
- \(-32t\) corresponds to the constant acceleration due to gravity, which decelerates the baseball at 32 feet per second every second.
- \(75\) is the baseball's initial velocity upward.
It's vital to note that when \( v(t) = 0 \), it indicates that for that moment, the ball has stopped moving upwards and is about to descend.
This "stopping" point happens when the equation \(-32t + 75 = 0\) is solved for \( t \), indicating when the ball reaches its peak height.

Quadratic Equation

A quadratic equation is a second-order polynomial equation commonly used in describing the height of projectiles in motion. The general form of a quadratic equation is

• \( ax^2 + bx + c = 0 \)

For our projectile, the height function is

• \( h(t) = -16t^2 + 75t + 30 \)

This fits perfectly into the quadratic form where

• \( a = -16 \)
• \( b = 75 \)
• \( c = 30 \)

In projectile motion problems, the quadratic equation becomes essential when calculating specific points like when the ball hits the ground (when \( h(t) = 0 \)). The standard method to solve for \( t \) when \( h(t) = 0 \) is using the quadratic formula:

• \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

By inputting the coefficients \( a, b, \) and \( c \), we can solve for \( t \), discarding the negative solution, which doesn't make sense in the context of time.

Maximum Height

The maximum height of a projectile is a key point in projectile motion. It represents the peak of the object's vertical ascent before it begins its descent. Here, we achieve the maximum height when the velocity of the baseball is zero, as this signals the transition from moving upward to falling downward.
To find out when this occurs, we solve

• \( v(t) = 0 \)

This simplifies to

• \(-32t + 75 = 0\)

Solving for \( t \) yields

• \( t = \frac{75}{32} \)

seconds.
This value of \( t \) tells us when the ball reaches its peak height. To find the height at this precise moment, substitute \( t = \frac{75}{32} \) back into the height function \( h(t) \):

• \( h\left(\frac{75}{32}\right) = -16 \left(\frac{75}{32}\right)^2 + 75\left(\frac{75}{32}\right) + 30 \)

Solving this calculation will give us the exact height at its maximum point. Understanding how to determine maximum height is essential for solving numerous real-world problems involving projectiles.