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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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Most popular questions from this chapter

The position of an object as a function of time is given as \(x=A t^{3}+B t^{2}+C t+D .\) The constants are \(A=2.1 \mathrm{~m} / \mathrm{s}^{3}\) \(B=1.0 \mathrm{~m} / \mathrm{s}^{2}, C=-4.1 \mathrm{~m} / \mathrm{s},\) and \(D=3 \mathrm{~m}\) a) What is the velocity of the object at \(t=10.0 \mathrm{~s}\) ? b) At what time(s) is the object at rest? c) What is the acceleration of the object at \(t=0.50 \mathrm{~s} ?\) d) Plot the acceleration as a function of time for the time interval from \(t=-10.0 \mathrm{~s}\) to \(t=10.0 \mathrm{~s}\).

A particle starts from rest at \(x=0\) and moves for \(20 \mathrm{~s}\) with an acceleration of \(+2.0 \mathrm{~cm} / \mathrm{s}^{2}\). For the next \(40 \mathrm{~s}\), the acceleration of the particle is \(-4.0 \mathrm{~cm} / \mathrm{s}^{2} .\) What is the position of the particle at the end of this motion?

You drop a water balloon straight down from your dormitory window \(80.0 \mathrm{~m}\) above your friend's head. At \(2.00 \mathrm{~s}\) after you drop the balloon, not realizing it has water in it your friend fires a dart from a gun, which is at the same height as his head, directly upward toward the balloon with an initial velocity of \(20.0 \mathrm{~m} / \mathrm{s}\). a) How long after you drop the balloon will the dart burst the balloon? b) How long after the dart hits the balloon will your friend have to move out of the way of the falling water? Assume the balloon breaks instantaneously at the touch of the dart.

How much time does it take for a car to accelerate from a standing start to \(22.2 \mathrm{~m} / \mathrm{s}\) if the acceleration is constant and the car covers \(243 \mathrm{~m}\) during the acceleration?

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