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Chapter 2: Problem 77

An object is thrown upward with a speed of \(28.0 \mathrm{~m} / \mathrm{s}\). How long does it take it to reach its maximum height?

Short Answer

Expert verified
Answer: The time it takes for the object to reach its maximum height is approximately 2.86 seconds.

Step by step solution

01

Identify the given values and the equation to be used

We are given the initial velocity of the object, \(v_0 = 28.0 \mathrm{~m} / \mathrm{s}\). We want to find the time \(t\) it takes to reach maximum height. At the maximum height, the object's final velocity \(v\) is 0 m/s. We can use the equation of motion that relates velocities and time, which is given by: \(v = v_0 - gt\) Where \(g\) is the acceleration due to gravity (\(9.81 \mathrm{~m} / \mathrm{s}^2\)).
02

Plug in the given values into the equation and solve for t

Now, we will substitute the given values into the equation: \(0 = 28.0 \mathrm{~m} / \mathrm{s} - 9.81 \mathrm{~m} / \mathrm{s}^2 \cdot t\) To solve for \(t\), we will first isolate the term with \(t\) on one side of the equation: \(9.81 \mathrm{~m} / \mathrm{s}^2 \cdot t = 28.0 \mathrm{~m} / \mathrm{s}\) Now, we will divide both sides of the equation by \(9.81 \mathrm{~m} / \mathrm{s}^2\) to find the value of \(t\): \(t = \frac{28.0 \mathrm{~m} / \mathrm{s}}{9.81 \mathrm{~m} / \mathrm{s}^2}\)
03

Calculate the time t and give the final answer

By dividing the numbers, we find the value of \(t\): \(t = \frac{28.0}{9.81} \approx 2.86 \mathrm{~s}\) It takes approximately \(2.86\) seconds for the object to reach its maximum height.

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Most popular questions from this chapter

In 2005, Hurricane Rita hit several states in the southern United States. In the panic to escape her wrath, thousands of people tried to flee Houston, Texas by car. One car full of college students traveling to Tyler, Texas, 199 miles north of Houston, moved at an average speed of \(3.0 \mathrm{~m} / \mathrm{s}\) for one-fourth of the time, then at \(4.5 \mathrm{~m} / \mathrm{s}\) for another one-fourth of the time, and at \(6.0 \mathrm{~m} / \mathrm{s}\) for the remainder of the trip. a) How long did it take the students to reach their destination? b) Sketch a graph of position versus time for the trip.

You ride your bike along a straight line from your house to a store \(1000 .\) m away. On your way back, you stop at a friend's house which is halfway between your house and the store. a) What is your displacement? b) What is the total distance traveled? After talking to your friend, you continue to your house. When you arrive back at your house, c) What is your displacement? d) What is the distance you have traveled?

Two train cars are on a straight, horizontal track. One car starts at rest and is put in motion with a constant acceleration of \(2.00 \mathrm{~m} / \mathrm{s}^{2}\). This car moves toward a second car that is \(30.0 \mathrm{~m}\) away and moving at a constant speed of \(4.00 \mathrm{~m} / \mathrm{s}\). a) Where will the cars collide? b) How long will it take for the cars to collide?

Starting from rest, a boat increases its speed to \(5.00 \mathrm{~m} / \mathrm{s}\) with constant acceleration. a) What is the boat's average speed? b) If it takes the boat 4.00 s to reach this speed, how far has it traveled?

Two cars are traveling at the same speed, and the drivers hit the brakes at the same time. The deceleration of one car is double that of the other. By what factor does the time required for that car to come to a stop compare with that for the other car?
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