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Chapter 2: Problem 70

A balloon is rising at \(10 \mathrm{m} / \mathrm{s}\) when its passenger throws a ball straight up at \(12 \mathrm{m} / \mathrm{s}\) relative to the balloon. How much later does the passenger catch the ball?

Short Answer

Expert verified
The passenger catches the ball 4.48 seconds later.

Step by step solution

01

Understand the context and identify the variables

First, understand that the balloon itself is rising at \(10 \mathrm{m} / \mathrm{s}\). Then when the ball is thrown upwards at \(12 \mathrm{m} / \mathrm{s}\), the resulting speed of the ball relative to the ground is \(10 \mathrm{m} / \mathrm{s} + 12 \mathrm{m} / \mathrm{s} = 22 \mathrm{m} / \mathrm{s}\). Then, we know that the velocity of the ball will eventually decrease due to gravity, which pulls it downwards at \(9.81 \mathrm{m} / \mathrm{s}^2\).
02

Find the time for the ball to reach its highest point

We need to know when the ball reaches its highest point. At that moment, the velocity of the ball relative to the ground will be zero. We can use the equation: final velocity = initial velocity + (acceleration \cdot time). We solve for time with this setup: \(0 = 22 \mathrm{m/s} + (-9.81 \mathrm{m/s^2} \cdot t)\). Solving this equation, we find that \(t = 22 / 9.81 = 2.24 \mathrm{s}\).
03

Calculate the total time taken

At the peak point, the ball will start falling back to the ground, but remember that the balloon is still rising. Therefore, the time the ball takes to reach back to the balloon after reaching its highest point is the same as the time taken to reach the highest point. The total time is twice the time taken to reach the highest point: \(2 \cdot 2.24 = 4.48 \mathrm{s}\).

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Most popular questions from this chapter

An object's position is given by \(x=b t^{3},\) with \(x\) in meters, \(t\) in seconds, and \(b=1.5 \mathrm{m} / \mathrm{s}^{3} .\) Determine (a) the instantaneous velocity and (b) the instantaneous acceleration at the end of \(2.5 \mathrm{s}\) Find (c) the average velocity and (d) the average acceleration during the first \(2.5 \mathrm{s}\)

You're speeding at \(85 \mathrm{km} / \mathrm{h}\) when you notice that you're only \(10 \mathrm{m}\) behind the car in front of you, which is moving at the legal speed limit of \(60 \mathrm{km} / \mathrm{h} .\) You slam on your brakes, and your car negatively accelerates at \(4.2 \mathrm{m} / \mathrm{s}^{2} .\) Assuming the other car continues at constant speed, will you collide? If so, at what relative speed? If not, what will be the distance between the cars at their closest approach?

A base runner can get from first to second base in 3.4 s. If he leaves first as the pitcher throws a 90 milh fastball the 61 -ft distance to the catcher, and if the catcher takes 0.45 s to catch and rethrow the ball, how fast does the catcher have to throw the ball to second base to make an out? Home plate to second base is the diagonal of a square \(90 \mathrm{ft}\) on a side.

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Is it possible to be at position \(x=0\) and still be moving?
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