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Chapter 2: Problem 78

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?

Short Answer

Expert verified
Answer: The object is 23.095 m above the projection point.

Step by step solution

01

Identify the parameters

Given the initial velocity \(v_0 = 28.0 \mathrm{~m/s}\) and the time \(t = 1.00 \mathrm{~s}\). The acceleration due to gravity is \(g = 9.81 \mathrm{~m/s^2}\), and it acts opposite to the initial velocity.
02

Choose the kinematic equation

We will use the following kinematic equation, which relates displacement, initial velocity, time, and acceleration: \(y = v_0t - \frac{1}{2}gt^2\). In this equation, \(y\) is the vertical displacement or height above the projection point.
03

Plug in the values and calculate the height

Insert the given values into the equation: \(y = (28.0 \mathrm{~m/s})(1.00 \mathrm{~s}) - \frac{1}{2}(9.81 \mathrm{~m/s^2})(1.00 \mathrm{~s})^2\). Now, perform the calculations: \(y = 28.0 \mathrm{~m} - 4.905 \mathrm{~m}\) \(y = 23.095 \mathrm{~m}\)
04

Report the final answer

The object is \(23.095 \mathrm{~m}\) above the projection point after \(1.00 \mathrm{~s}\).

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Most popular questions from this chapter

You drive with a constant speed of \(13.5 \mathrm{~m} / \mathrm{s}\) for 30.0 s. You then accelerate for 10.0 s to a speed of \(22.0 \mathrm{~m} / \mathrm{s}\). You then slow to a stop in \(10.0 \mathrm{~s}\). How far have you traveled?

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