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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

You are driving at \(29.1 \mathrm{~m} / \mathrm{s}\) when the truck ahead of you comes to a halt \(200.0 \mathrm{~m}\) away from your bumper. Your brakes are in poor condition and you decelerate at a constant rate of \(2.4 \mathrm{~m} / \mathrm{s}^{2}\) a) How close do you come to the bumper of the truck? b) How long does it take you to come to a stop?

You are trying to improve your shooting skills by shooting at a can on top of a fence post. You miss the can, and the bullet, moving at \(200 . \mathrm{m} / \mathrm{s},\) is embedded \(1.5 \mathrm{~cm}\) into the post when it comes to a stop. If constant acceleration is assumed, how long does it take for the bullet to stop?

A car moving at \(60.0 \mathrm{~km} / \mathrm{h}\) comes to a stop in \(t=4.00 \mathrm{~s}\) Assume uniform deceleration. a) How far does the car travel while stopping? b) What is its deceleration?

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?

A car travels \(22.0 \mathrm{~m} / \mathrm{s}\) north for \(30.0 \mathrm{~min}\) and then reverses direction and travels \(28.0 \mathrm{~m} / \mathrm{s}\) for \(15.0 \mathrm{~min}\). What is the car's total displacement? a) \(1.44 \cdot 10^{4} \mathrm{~m}\) b) \(6.48 \cdot 10^{4} \mathrm{~m}\) c) \(3.96 \cdot 10^{4} \mathrm{~m}\) d) \(9.98 \cdot 10^{4} \mathrm{~m}\)
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