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Chapter 3: Problem 62

A ball is dropped from the top of the Empire State Building. The height, \(y,\) of the ball above the ground (in feet) is given as a function of time, \(t,\) (in seconds) by $$y=1250-16 t^{2}$$ (a) Find the velocity of the ball at time \(t .\) What is the sign of the velocity? Why is this to be expected? (b) When does the ball hit the ground, and how fast is it going at that time? Give your answer in feet per second and in miles per hour \((1 \mathrm{ft} / \mathrm{sec}=15 / 22 \mathrm{mph})\)

Short Answer

Expert verified
(a) \(v(t) = -32t\), negative (downward). (b) Hits at \(8.84\) sec, \(-282.88\) ft/sec or \(192.72\) mph.

Step by step solution

01

Find the velocity function

The velocity of the ball as a function of time is the derivative of the height function with respect to time. Hence, take the derivative of \( y = 1250 - 16t^2 \). Using the power rule, the derivative is \( v(t) = \frac{dy}{dt} = -32t \).
02

Determine the sign of the velocity

The velocity function is \( v(t) = -32t \). Since time \( t \geq 0 \), the velocity will always be negative. This negative sign indicates that the ball is moving downward.
03

Find when the ball hits the ground

The ball hits the ground when \( y = 0 \). Set \( 1250 - 16t^2 = 0 \) and solve for \( t \). Rearrange to \( 16t^2 = 1250 \), yielding \( t^2 = \frac{1250}{16} = 78.125 \). Taking the square root gives \( t \approx 8.84 \) seconds.
04

Calculate the velocity when the ball hits the ground

Substitute \( t = 8.84 \) into the velocity function: \( v(8.84) = -32(8.84) \). This gives \( v(8.84) \approx -282.88 \) feet per second.
05

Convert velocity to miles per hour

Use the conversion factor \( 1 \text{ ft/sec} = \frac{15}{22} \text{ mph} \). The velocity in miles per hour is \( \frac{15}{22} \times 282.88 \approx 192.72 \) mph.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Derivative
In the context of calculus, the derivative represents the rate at which one quantity changes with respect to another. In this exercise, we have a function describing the height of a ball over time: \(y = 1250 - 16t^2\). The derivative of this function with respect to time, \(t\), is the velocity of the ball. The derivative is found using simple calculus techniques, particularly the power rule.
  • The power rule states that if you have a term \(at^n\), its derivative will be \(nat^{n-1}\).
  • Applying the rule here: the derivative of \(-16t^2\) is \(-32t\), because you multiply the exponent by the coefficient and reduce the exponent by one.
  • The derivative of a constant, like 1250, is zero.
Therefore, the rate of change of height with respect to time, or the velocity, is \(v(t) = -32t\). This derivative helps us understand how fast and in which direction the ball is moving at any given time.
Velocity
Velocity is a key concept in understanding motion. It describes the speed and direction of an object. Unlike speed, which is always positive, velocity can be negative, indicating direction.
  • In this problem, the velocity is given by the function \(v(t) = -32t\).
  • The negative sign in front of the velocity function indicates that the ball is moving downward.
  • As \(t\) increases, the magnitude of the velocity increases, reflecting that the ball is speeding up as it falls.
Velocity can tell us more than just speed; it informs us about the direction of motion. Since the ball is dropped, it is reasonable for the velocity to be negative, showing it is moving towards the ground. Additionally, at the moment the ball hits the ground, the velocity indicates how fast it is moving. Converting the velocity from feet per second to miles per hour adds a more intuitive understanding of just how fast that is in everyday terms.
Motion
Motion is the change of position of an object over time, and it can be described quantitatively using physics and calculus. In this exercise, the motion of the ball is modeled by its height as a function of time: \(y = 1250 - 16t^2\). Key aspects of this motion can be explored by analyzing the function.
  • The height function is quadratic, indicating uniform acceleration due to gravity, hence the term \(-16t^2\).
  • Gravity acts on the ball from the moment it is released, causing it to accelerate downward, increasing its speed until it hits the ground.
  • The ball's motion continues until it reaches the ground, which is determined by setting \(y = 0\) and solving for \(t\), resulting in approximately 8.84 seconds.
Understanding motion involves not just where and when something is, but also how its velocity and position change over time. As time progresses, the ball moves closer to the ground until it finally hits and stops, manifesting the concepts of calculus in real-world motion.

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