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Chapter 2: Problem 58

A ball is tossed vertically upward with an initial speed of \(26.4 \mathrm{~m} / \mathrm{s}\). How long does it take before the ball is back on the ground?

Short Answer

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Question: A ball is thrown vertically upward with an initial speed of 26.4 m/s. Calculate the total time taken by the ball to be back on the ground. Answer: The ball takes 5.38 seconds before it is back on the ground.

Step by step solution

01

Identify the given parameters and unknowns

We are given: Initial velocity (u) = \(26.4 \mathrm{~m/s}\) Final velocity at maximum height (v) = \(0 \mathrm{~m/s}\) //Since the ball will stop momentarily at the highest point Acceleration due to gravity (a) = \(-9.81 \mathrm{~m/s^2}\) // The gravitational force acts in the downward direction, opposite to the direction of the initial velocity, hence it is negative We need to find the time (t) it takes for the ball to be back on the ground.
02

Calculate the time taken to reach maximum height

Using the first equation of motion: $$v = u + at$$ We can solve for the time taken to reach maximum height (t1), when the ball stops momentarily (v = 0). $$0 = 26.4 + (-9.81) t1$$ Now, solve for t1: $$t1 = \frac{26.4}{9.81} = 2.69 \mathrm{~s}$$
03

Calculate the time taken to fall back to the ground

Since it's a symmetrical situation, the time taken for the ball to fall back to the ground will be the same as the time taken to reach its maximum height. So, the time taken to fall back to the ground (t2) will be the same as t1. Thus, $$t2 = 2.69 \mathrm{~s}$$
04

Calculate the total time before the ball is back on the ground

Now we can calculate the total time (t) by adding the time taken to reach maximum height (t1) and the time taken to fall back to the ground (t2): $$t = t1 + t2$$ $$t = 2.69 + 2.69 = 5.38 \mathrm{~s}$$ So, it takes the ball \(5.38 \mathrm{~s}\) before it is back on the ground.

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