Problem 72 - Calculus and its applications

Question

A football player punts a football, which leaves his foot at a height of \(3 \mathrm{ft}\) above the ground and with an initial upward velocity of \(70 \mathrm{ft} /\) sec. The vertical velocity of the football \(t\) seconds after it is punted is given by $$v(t)=-32 t+70$$ where \(v\) is in feet per second. a) Find the function \(h\) that gives the height (in feet) of the football after \(t\) seconds. b) What are the height and the velocity of the football after 1.5 sec? c) After how many seconds does the ball reach its highest point, and how high is the ball at this point? d) The punt returner catches the football \(5 \mathrm{ft}\) above the ground. What is the vertical velocity of the ball immediately before it is caught?

Step-by-Step Solution

Answer

a) \( h(t) = -16t^2 + 70t + 3 \). b) Height: 104 ft, Velocity: 22 ft/s at 1.5 sec. c) Maximum height: 80.98 ft at 2.19 sec. d) Velocity just before catching: 38.5 ft/s.

Step 1: Find the Expression for Height

To find the function for height, use the physics equation for motion: \( h(t) = -\frac{1}{2}gt^2 + v_0t + h_0 \), where \( g = 32 \ \text{ft/s}^2\) is the acceleration due to gravity, \( v_0 = 70 \ \text{ft/s} \) is the initial upward velocity, and \( h_0 = 3 \ \text{ft} \) is the initial height. Plug in these values to get:\[ h(t) = -16t^2 + 70t + 3. \]

Step 2: Substitute to Find Height and Velocity at 1.5 Seconds

Substitute \( t = 1.5 \) into the expression for height: \[ h(1.5) = -16(1.5)^2 + 70(1.5) + 3. \]Calculate this to find the height. Next, use the velocity function \( v(t) = -32t + 70 \) and substitute \( t = 1.5 \), \[ v(1.5) = -32(1.5) + 70. \]Calculate this to find the velocity.

Step 3: Determine the Time and Height at the Highest Point

The maximum height is reached when vertical velocity is zero. Set \( v(t) = 0 \) to solve for \( t \):\[ 0 = -32t + 70 \] \[ 32t = 70 \] \[ t = \frac{70}{32} = 2.1875 \text{ seconds.} \]Substitute \( t = 2.1875 \) into the height equation to find the maximum height:\[ h(2.1875) = -16(2.1875)^2 + 70(2.1875) + 3. \]

Step 4: Find Velocity Just Before the Football is Caught

To find the velocity when the football is 5 feet above the ground, solve for \( t \) when \( h(t) = 5 \). Use the quadratic equation:\[ -16t^2 + 70t + 3 = 5. \]Simplify and solve for \( t \):\[ -16t^2 + 70t - 2 = 0. \]Use the quadratic formula, \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), with \( a = -16 \), \( b = 70 \), \( c = -2 \). Calculate the positive root for \( t \). Substitute this \( t \) value into \( v(t) = -32t + 70 \) to find the velocity just before the football is caught.

Key Concepts

Quadratic Equations and Their Role in Projectiles

Quadratic equations often appear in physics, especially when dealing with projectile motion. When a football is kicked, its path follows a curved trajectory. This trajectory can be described using quadratic equations. In general, a quadratic equation has the form \( ax^2 + bx + c = 0 \). When applied to projectiles, the equation describes how the height \( h \) of an object changes over time \( t \).
The equation given is \( h(t) = -16t^2 + 70t + 3 \), showcasing the principles of quadratic equations.

• The term \(-16t^2\) represents the effect of gravity, pulling the football downwards.
• The term \(70t\) reflects the initial upward velocity that propels the football into the air.
• The constant term \(3\) is the initial position of the ball, indicating it was kicked from 3 feet above ground.

Understanding the elements of quadratic equations helps in predicting and analyzing projectile paths, allowing us to determine critical points like maximum height or when the projectile will reach a certain level.

Understanding Velocity in Projectile Motion

Velocity is a vector quantity, meaning it has both magnitude and direction. In the context of projectile motion, velocity is dynamic and changes over time due to gravitational forces.

The problem involves understanding the equation \( v(t) = -32t + 70 \), which offers an insight into how the football's vertical velocity changes:

• The term \(-32t\) captures the constant deceleration caused by gravity, affecting the football's upward motion.
• The term \(70\) signifies the initial velocity, or the speed at which the ball leaves the player's foot.

At different time points, you can rearrange and substitute \( t \) to find the specific vertical velocity at that moment.
For instance, setting \( v(t) = 0 \) helps identify when the ball reaches its peak height, as that's when it stops ascending but hasn't started descending.

Physics Equations in Motion Analysis

Physics equations allow us to predict motion and behavior of moving objects like the football in question. They integrate different physical principles that govern how objects move.

The primary physics equation used here is :
\[ h(t) = -\frac{1}{2}gt^2 + v_0t + h_0 \]
This equation incorporates gravitational acceleration \( g = 32 \ \text{ft/s}^2 \), initial velocity \( v_0 = 70 \ \text{ft/s} \), and initial height \( h_0 = 3 \ \text{ft} \).

• Gravitational acceleration is a constant that acts to pull objects back to earth.
• Initial velocity and height determine the starting conditions of the projectile motion.

This combination of physics equations and projectile properties helps compute essential details, such as the time taken to reach maximum height, the highest point reached, and when the object returns to a specific position, such as 5 feet above the ground in this case.