Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 2: Problem 62

An object is thrown vertically upward and has a speed of \(20.0 \mathrm{~m} / \mathrm{s}\) when it reaches two thirds of its maximum height above the launch point. Determine its maximum height.

Short Answer

Expert verified
Answer: The maximum height reached by the object is approximately 40.61 m.

Step by step solution

01

Identify the known variables

We know the following: - Speed at two thirds of maximum height, \(v = 20.0\mathrm{~m/s}\) - Fraction of the maximum height, \(h' = \dfrac{2}{3} H\) - Acceleration due to gravity, \(a = -9.81\mathrm{~m/s^2}\) (downward, hence negative) - Final velocity at maximum height, \(v_f = 0\mathrm{~m/s^2}\) Our goal is to find the maximum height, \(H\).
02

Use the equation of motion to find the initial velocity at two thirds height

The equation of motion that relates the velocity, the initial velocity, the height, and the acceleration is: \(v^2 = v_i^2 + 2a(h - h_i)\). At two thirds of the maximum height, the above equation becomes: \(v^2 = v_i^2 + 2a(h' - h_i)\). Since the object is thrown from the ground, we know that \(h_i = 0\). So, we have: \(v^2 = v_i^2 + 2a h'\). Solve for \(v_i\) using the given values: \((20.0\mathrm{~m/s})^2 = v_i^2 + 2(-9.81\mathrm{~m/s^2})(\dfrac{2}{3}H)\).
03

Determine initial velocity at the ground

Rearrange the equation above to find the initial velocity: \(v_i^2 = (20.0\mathrm{~m/s})^2 - 2(-9.81\mathrm{~m/s^2})(\dfrac{2}{3}H)\).
04

Calculate the final velocity at maximum height

Since the object reaches the maximum height and momentarily comes to rest, we know its final velocity, \(v_f = 0\mathrm{~m/s}\). Now we can use another equation of motion to relate the initial and final velocities, height, and acceleration: \(v_f^2 = v_i^2 + 2aH\).
05

Plug in the initial velocity equation to find the maximum height

Substitute the equation for \(v_i^2\) found in step 3 into the equation of motion in step 4: \(0\mathrm{~m/s^2} = ((20.0\mathrm{~m/s})^2 - 2(-9.81\mathrm{~m/s^2})(\dfrac{2}{3}H))+ 2(-9.81\mathrm{~m/s^2})H\).
06

Solve for the maximum height H

Now we can solve for \(H\). First, simplify the equation: \(0 = (20.0\mathrm{~m/s})^2 - (\dfrac{4}{3})(-9.81\mathrm{~m/s^2})H + 2(-9.81\mathrm{~m/s^2})H\). Next, rearrange and solve for \(H\): \(H = \dfrac{(20.0\mathrm{~m/s})^2}{(\dfrac{-2}{3})(-9.81\mathrm{~m/s^2})}\). \(H \approx 40.61\mathrm{~m}\). The maximum height reached by the object is approximately \(40.61\mathrm{~m}\).

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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

An object starts from rest and has an acceleration given by \(a=B t^{2}-\frac{1}{2} C t,\) where \(B=2.0 \mathrm{~m} / \mathrm{s}^{4}\) and \(C=-4.0 \mathrm{~m} / \mathrm{s}^{3}\). a) What is the object's velocity after 5.0 s? b) How far has the object moved after \(t=5.0\) s?

In a fancy hotel, the back of the elevator is made of glass so that you can enjoy a lovely view on your ride. The elevator travels at an average speed of \(1.75 \mathrm{~m} / \mathrm{s}\). A boy on the 15th floor, \(80.0 \mathrm{~m}\) above the ground level, drops a rock at the same instant the elevator starts its ascent from the 1st to the 5th floor. Assume the elevator travels at its average speed for the entire trip and neglect the dimensions of the elevator. a) How long after it was dropped do you see the rock? b) How long does it take for the rock to reach ground level?

An object is thrown upward with a speed of \(28.0 \mathrm{~m} / \mathrm{s}\). How high above the projection point is it after 1.00 s?

The position of a race car on a straight track is given as \(x=a t^{3}+b t^{2}+c,\) where \(a=2.0 \mathrm{~m} / \mathrm{s}^{3}, b=2.0 \mathrm{~m} / \mathrm{s}^{2}\), and \(c=3.0 \mathrm{~m}\). a) What is the car's position between \(t=4.0 \mathrm{~s}\) and \(t=9.0 \mathrm{~s}\) ? b) What is the average speed between \(t=4.0 \mathrm{~s}\) and \(t=9.0 \mathrm{~s} ?\)

You drop a water balloon straight down from your dormitory window \(80.0 \mathrm{~m}\) above your friend's head. At \(2.00 \mathrm{~s}\) after you drop the balloon, not realizing it has water in it your friend fires a dart from a gun, which is at the same height as his head, directly upward toward the balloon with an initial velocity of \(20.0 \mathrm{~m} / \mathrm{s}\). a) How long after you drop the balloon will the dart burst the balloon? b) How long after the dart hits the balloon will your friend have to move out of the way of the falling water? Assume the balloon breaks instantaneously at the touch of the dart.
See all solutions

Recommended explanations on Physics Textbooks

Classical Mechanics

Read Explanation

Turning Points in Physics

Read Explanation

Physics of Motion

Read Explanation

Energy Physics

Read Explanation

Fundamentals of Physics

Read Explanation

Famous Physicists

Read Explanation
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.

Sign-up for free