/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 50 A rock dropped from a 1024-foot-... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A rock dropped from a 1024-foot-high cliff falls a distance \(D\) given by \(D=16 t^2\), where \(t\) is the time in seconds after the rock is dropped. How long will it take the rock to reach the bottom of the cliff? A. 64 seconds B. 32 seconds C. 16 seconds D. 8 seconds

Short Answer

Expert verified
D. 8 seconds

Step by step solution

01

Identify the given information

The cliff is 1024 feet high. The distance the rock falls, \(D\), is given by the equation \(D = 16t^2\).
02

Set up the equation

Since the rock is falling from a height of 1024 feet, we set \(D\) to 1024. So, the equation becomes \(1024 = 16t^2\).
03

Solve for \(t^2\)

Divide both sides of the equation by 16 to isolate \(t^2\): \(\frac{1024}{16} = t^2\), which simplifies to \(64 = t^2\).
04

Take the square root

Take the square root of both sides to solve for \(t\): \(t = \sqrt{64}\). This results in \(t = 8\).
05

Verify the correct answer choice

The time it takes for the rock to reach the bottom of the cliff is 8 seconds. Therefore, the correct answer is D. 8 seconds.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

physics problems
Physics problems can often be a bit tricky but exciting! They help us understand the laws of nature. In this exercise, we are dealing with the motion of a rock falling from a cliff. This type of motion is called 'free fall' and is heavily influenced by gravity. When an object is in free fall, it accelerates downwards at a constant rate due to gravity. The formula used here, \(D = 16t^2\), combines this acceleration and the time to tell us how far the rock will travel in a given time. The number 16 here actually comes from the simplification of the gravitational acceleration on Earth (\textrm{32 ft/s²} divided by 2). This makes it easier to predict the distance an object will fall.
distance and time equations
Equations that relate distance and time are fundamental in both math and physics. They allow us to calculate one quantity if we know the other. In our case, the equation \(D = 16t^2\) relates the distance \(D\) fallen by the rock to the time that has passed since it was dropped. To solve the problem, we substituted the known distance (1024 feet) into the equation and solved for time (\(t\)). Here’s how it worked:
  • Identify the given values and equation: 1024 feet and \(D = 16t^2\).
  • Substitute the distance: \(1024 = 16t^2\).
  • Isolate \(t^2\): \(\frac{1024}{16} = t^2\), which simplifies to \(64 = t^2\).
This leaves you with a simple last step involving square roots.
square root calculation
Taking square roots is a necessary skill in many areas of math and science. It helps us find the original value before it was squared. To isolate the time \(t\) in our problem, we used the square root. Here's how:
  • We had the equation \(64 = t^2\).
  • Taking the square root of both sides, we get \(t = \sqrt{64}\).
  • Simplifying \sqrt{64}, we find \(t = 8\).
Always remember, the square root of a number is a value that, when multiplied by itself, gives the original number. In our case, \(\frac{64}{8} = 8\), confirming that \(t = 8\) seconds is indeed correct. This final step is crucial because it converts mathematical answers into real-world time units, clarifying that the rock takes 8 seconds to hit the ground.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Solve \(p-5=12\)

Aaron is earning money by mowing lawns. He charges \(\$ 40\) per lawn. He is trying to save up at least \(\$ 1500\) for a new riding lawnmower. \(\mathrm{He}\) already has \(\$ 420\). Which inequality best represents Aaron's riding lawnmower goal? A. \(420 x+40 \geq 1500\) B. \(40 x+420 \geq 1500\) C. \(420 x+40 \leq 1500\) D. \(40 x+420 \leq 1500\)

Constance is controlling the flow of a solution into a graduated cylinder. She wants to go to lunch and decides to set the flow at a lower rate rather than shut it off completely. The cylinder already holds 72 milliliters of solution and can hold a maximum of 500 milliliters. If \(r\) represents the rate at which the flow is set in milliliters per minute, which inequality could Constance solve to determine a safe range of flow rates, assuming she plans to take 60 minutes for lunch? A. \(72 r+60 \geq 500\) B. \(72 r+60 \leq 500\) C. \(60 r+72 \geq 500\) D. \(60 r+72 \leq 500\)

A passenger train leaves the station, traveling west at 120 miles per hour. A freight train leaves the same station 2 hours later traveling 80 miles per hour. If \(t\) represents the time in hours that the first train has traveled, which equation represents a situation where the two trains are 640 miles apart? A. \(120 t+80(t-2)=640\) B. \(120(t+2)+80 t=640\) C. \(120 t+80(t+2)=640\) D. \(120(t-2)+80 t=640\)

Vicki is selling bags of caramel corn at her booth at the state fair at a rate of 5 bags per hour. By noon, she had sold 15 bags. She will have sold a total of 40 bags at \begin{tabular}{|l|l|} \hline Select... & \(\nabla\) \\ \hline \(4: 00 \mathrm{PM}\) & \\ \hline \(5: 00 \mathrm{PM}\) \\ \hline 6:00 PM \\ \hline \end{tabular}

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.