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Chapter 2: Problem 41

(II) A ball player catches a ball 3.4 s after throwing it vertically upward. With what speed did he throw it, and what height did it reach?

Short Answer

Expert verified
The ball was thrown at 16.66 m/s and reached a height of 14.16 m.

Step by step solution

01

Understand the Problem

A ball is thrown vertically upward and is caught after 3.4 seconds. We need to find the initial speed at which the ball was thrown and the maximum height it reached.
02

Recall the Equations of Motion

The motion of the ball is governed by two main kinematic equations: 1. To find the initial speed: \( v = u + at \), where \( v \) is the final velocity, \( u \) is the initial velocity, \( a = -9.8 \, m/s^2 \) (acceleration due to gravity), and \( t = 3.4 \, s \).2. To find the maximum height: \( h = ut + \frac{1}{2}at^2 \).
03

Find the Time to Reach Maximum Height

The total time for the round trip of the ball is 3.4 seconds, which means it takes half of this time to reach the maximum height. Thus, \( t_{up} = \frac{3.4}{2} = 1.7 \, s \).
04

Calculate Initial Speed

Using the first equation of motion, where the final velocity (\( v = 0 \)) at maximum height, we set up the equation: 0 = u - 9.8 \times 1.7. Solving for \( u \), we find \( u = 9.8 \times 1.7 = 16.66 \, m/s \).
05

Calculate Maximum Height

Using the second equation of motion: \( h = ut_{up} + \frac{1}{2}at_{up}^2 \), we substitute \( u = 16.66 \, m/s\), \( a = -9.8 \, m/s^2 \), and \( t_{up} = 1.7\) to get: \[h = 16.66 \times 1.7 + \frac{1}{2} \times (-9.8) \times (1.7)^2 \]Calculating, \( h = 28.322 - 14.161 = 14.161 \, m \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Equations of Motion
The equations of motion are fundamental to understanding the behavior of objects in kinematics. These equations relate the five key variables of motion: displacement, initial velocity, final velocity, acceleration, and time. In the given problem, we use these equations to find the initial velocity and the maximum height reached by the ball.

  • For finding the initial velocity, the equation used is: \( v = u + at \), where \( v \) is the final velocity, \( u \) is the initial velocity, \( a \) is acceleration, and \( t \) is time. In this problem, the final velocity becomes zero at the highest point of the throw.

  • To find the maximum height: \( h = ut + \frac{1}{2}at^2 \), where \( h \) is the height, \( u \) is the initial velocity, \( a \) is acceleration and \( t \) is the time to reach maximum height.

These equations help decode each element of the ball's motion, letting us both predict and analyze physical scenarios involving motion along a straight line.
Acceleration Due to Gravity
The acceleration due to gravity is a constant that affects every object moving near the Earth's surface. This force pulls the object towards the ground at a rate of \(-9.8 \, m/s^2\).

  • It's negative because it acts downward, opposite to the direction in which the ball was initially thrown.

  • In this problem, gravity acts to decelerate the upward motion of the ball and subsequently accelerates it back down.

This acceleration impacts all falling objects equally, ignoring air resistance. Thus, when calculating the ball's trajectory, it's crucial to account for this constant force to predict how long the ball will stay in the air and where it will land.
Initial Velocity
The initial velocity is the speed at which an object begins its motion. For the ball in our problem, this is the speed at which it is thrown straight up into the air.

  • Using the equation of motion, we solve for the initial velocity. Since the final velocity at the peak is zero, the equation simplifies to finding the initial speed required to stop the ball at its maximum height.

  • From the solution, we find that the initial velocity, \( u \), is \( 16.66 \, m/s \).

The concept of initial velocity is crucial in determining the motion's trajectory, height, and duration. It sets the foundation for how far and how long the object will travel after being thrown.

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Most popular questions from this chapter

In putting, the force with which a golfer strikes a ball is planned so that the ball will stop within some small distance of the cup, say 1.0 m long or short, in case the putt is missed. Accomplishing this from an uphill lie (that is, putting the ball downhill, see Fig. 2-47) is more difficult than from a downhill lie. To see why, assume that on a particular green the ball decelerates constantly at 1.8 m/s\(^2\) going downhill, and constantly at 2.6 m/s\(^2\) going uphill. Suppose we have an uphill lie 7.0 m from the cup. Calculate the allowable range of initial velocities we may impart to the ball so that it stops in the range 1.0 m short to 1.0 m long of the cup. Do the same for a downhill lie 7.0 m from the cup. What in your results suggests that the downhill putt is more difficult?

If there were no air resistance, how long would it take a free-falling skydiver to fall from a plane at 3200 m to an altitude of 450 m, where she will open her parachute? What would her speed be at 450 m? (In reality, the air resistance will restrict her speed to perhaps 150 km/h.)

(II) A space vehicle accelerates uniformly from 85 m/s at \(t =\) 0 to 162 m/s at \(t =\) 10.0 s. How far did it move between \(t =\) 2.0 s and \(t =\) 6.0 s

(II) A car traveling at 95 km/h strikes a tree. The front end of the car compresses and the driver comes to rest after traveling 0.80 m. What was the magnitude of the average acceleration of the driver during the collision? Express the answer in terms of "\(g\)'s," where 1.00 \(g =\) 9.80 m/s\(^2\).

(I) A bird can fly 25 km/h. How long does it take to fly 3.5 km?
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