Problem 154 - Calculus Volume 1

Question

A ball is thrown downward with a speed of 8 \(\mathrm{ft} / \mathrm{s}\) from the top of a 64 -foot-tall building. After \(t\) seconds, its height above the ground is given by \(s(t)=-16 t^{2}-8 t+64\) a. Determine how long it takes for the ball to hit the ground. b. Determine the velocity of the ball when it hits the ground.

Step-by-Step Solution

Answer

a. 1.77 seconds; b. -56.64 ft/s.

Step 1: Set the height equation to zero

To find out when the ball hits the ground, we need the height over time equation to be equal to 0: \[-16t^2 - 8t + 64 = 0\].

Step 2: Solve the quadratic equation

We solve the equation \(-16t^2 - 8t + 64 = 0\) using the quadratic formula: \[t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]. Here, \(a = -16\), \(b = -8\), and \(c = 64\).

Step 3: Calculate the discriminant

The discriminant \(b^2 - 4ac\) is \((-8)^2 - 4(-16)(64) = 64 + 4096 = 4160\).

Step 4: Apply the quadratic formula

Substitute the values into the quadratic formula: \[t = \frac{-(-8) \pm \sqrt{4160}}{2(-16)} = \frac{8 \pm \sqrt{4160}}{-32}\].

Step 5: Evaluate the solutions

Calculate the square root of 4160, which is approximately 64.496. Then solve for \(t\): \[t = \frac{8 \pm 64.496}{-32}\]. This gives two potential values for \(t\): \[t = \frac{8 + 64.496}{-32} = -2.265\] and \[t = \frac{8 - 64.496}{-32} = 1.77\]. Since time cannot be negative, \(t = 1.77\) seconds is the solution.

Step 6: Determine initial velocity

The given initial velocity of the ball is \(8 \mathrm{ft/s}\) downward, and the motion is affected by gravity at \(-32 \mathrm{ft/s}^2\).

Step 7: Calculate the velocity when the ball hits the ground

The velocity as a function of time is given by the derivative of the height function: \[v(t) = \frac{d}{dt}(-16t^2 - 8t + 64) = -32t - 8\].

Step 8: Substitute the time into the velocity equation

Substitute \(t = 1.77\) into the velocity function: \[v(1.77) = -32(1.77) - 8 = -56.64\]. This means the velocity of the ball when it hits the ground is approximately \(-56.64 \mathrm{ft/s}\).

Key Concepts

Projectile motion

Projectile motion occurs when an object is thrown into the air under the influence of gravity, following a curved path in the shape of a parabola. This type of motion is driven by two key components:

• A vertical component, which is influenced by gravitational acceleration and affects how high or low the object goes.
• A horizontal component, which remains constant in a frictionless environment because gravity does not influence horizontal movement.

In our exercise, the ball is thrown downward, making the initial vertical speed important.
Gravity acts as a constant vertical acceleration at \(-32 \mathrm{ft/s^2}\), pulling the ball towards the ground. Understanding these components helps in unraveling the different stages of the ball’s journey until it hits the ground. The method outlined in the exercise shows how to utilize the quadratic equation to find out the time at which the projectile reaches ground level or height zero.

Kinematics

Kinematics encompasses the study of motion without considering its causes. It primarily deals with calculations involving time, velocity, acceleration, and displacement.
Kinematic equations help describe the motion of objects and can be applied to different scenarios, such as linear motion or projectile motion.
In the problem provided, the time it takes for the ball to hit the ground is determined by solving a quadratic equation derived from the kinematics equation:\[s(t) = -16t^2 - 8t + 64\]This equation already incorporates gravitational acceleration, which is evident in the \(-16t^2\) term.
Another significant aspect of kinematics highlighted in the solution is velocity, which in this scenario differs from speed because it has direction — downward in this case. By using derivative principles, a function for velocity can be derived from the original position-time equation, which then allows us to find the velocity at any time \(t\). This is critical for determining the speed of the ball at the moment it impacts the ground.

Derivative of a function

Derivatives serve as fundamental tools in calculus to help understand the rate at which one quantity changes with respect to another.
When applied to motion, derivatives can express how velocity or acceleration changes over time. In our case, the derivative of the height function, \(s(t) = -16t^2 - 8t + 64\), provides insight into the ball's velocity:\[v(t) = \frac{d}{dt}(-16t^2 - 8t + 64) = -32t - 8\]Here, the derivative shows that the velocity is not constant but changes linearly with time due to the constant acceleration of gravity pushing the ball downward.
By substituting the calculated time into this velocity function, we can precisely determine how fast the ball is traveling when it contacts the ground. This use of derivatives enables one to analyze changes not immediately apparent from the original equation, unlocking deeper insights into motion dynamics.