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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

The velocity as a function of time for a car on an amusement park ride is given as \(v=A t^{2}+B t\) with constants \(A=2.0 \mathrm{~m} / \mathrm{s}^{3}\) and \(B=1.0 \mathrm{~m} / \mathrm{s}^{2} .\) If the car starts at the origin, what is its position at \(t=3.0\) s?

A girl is standing at the edge of a cliff \(100 . \mathrm{m}\) above the ground. She reaches out over the edge of the cliff and throws a rock straight upward with a speed \(8.00 \mathrm{~m} / \mathrm{s}\). a) How long does it take the rock to hit the ground? b) What is the speed of the rock the instant before it hits the ground?

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An object is thrown upward with a speed of \(28.0 \mathrm{~m} / \mathrm{s}\). What maximum height above the projection point does it reach?

A ball is dropped from the roof of a building. It hits the ground and it is caught at its original height 5.0 s later. a) What was the speed of the ball just before it hits the ground? b) How tall was the building? c) You are watching from a window \(2.5 \mathrm{~m}\) above the ground. The window opening is \(1.2 \mathrm{~m}\) from the top to the bottom. At what time after the ball was dropped did you first see the ball in the window?
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