/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 14 A ball that is thrown vertically... [FREE SOLUTION] | 91Ó°ÊÓ

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A ball that is thrown vertically upward on the Moon returns to its starting point in \(4.0 \mathrm{~s}\). The acceleration due to gravity there is \(1.60\) \(\mathrm{m} / \mathrm{s}^{2}\) downward. Find the ball's original speed. Take up as positive. For the trip from beginning to end, \(y=0\) (it ends at the same level it started at), \(a=-1.60 \mathrm{~m} / \mathrm{s}^{2}, t=4.0 \mathrm{~s}\). Use \(y=v_{i y} t+\frac{1}{2} a t^{2}\) to find $$ 0=v_{i y}(4.0 \mathrm{~s})+\frac{1}{2}\left(-1.60 \mathrm{~m} / \mathrm{s}^{2}\right)(4.0 \mathrm{~s})^{2} $$ from which \(v_{i y}=3.2 \mathrm{~m} / \mathrm{s}\).

Short Answer

Expert verified
The ball's original speed is 3.2 m/s.

Step by step solution

01

Identify known values

We are given that the ball returns to the starting point in 4.0 seconds. Therefore, the total time of flight, \( t = 4.0 \) s. The acceleration due to gravity on the Moon is \( a = -1.60 \) m/s\(^2\) (negative since it is downward). The displacement \( y = 0 \) since the ball returns to its original height.
02

Use the motion equation

The equation of motion to use is \[ y = v_{i_y}t + \frac{1}{2}a t^2 \] where \( y \) is the displacement, \( v_{i_y} \) is the initial vertical velocity, \( a \) is the acceleration, and \( t \) is the time.
03

Substitute known values into the equation

Since the displacement \( y \) is 0, we substitute the known values into the equation: \[ 0 = v_{i_y} (4.0) + \frac{1}{2}(-1.60)(4.0)^2 \] which simplifies to: \[ 0 = 4.0v_{i_y} - 0.5 \times 1.60 \times 16 \] \[ 0 = 4.0v_{i_y} - 12.8 \]
04

Solve for initial velocity \( v_{i_y} \)

Rearrange the equation to solve for \( v_{i_y} \): \[ 4.0v_{i_y} = 12.8 \] Dividing both sides by 4.0 gives: \[ v_{i_y} = \frac{12.8}{4.0} = 3.2 \] m/s.

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Key Concepts

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

Acceleration due to Gravity
The acceleration due to gravity is a crucial concept when dealing with projectile motion problems, such as those where objects are thrown vertically. On the Moon, the acceleration due to gravity is significantly lower than on Earth. Specifically, it is about 1.60 m/s² downwards. This means that any object thrown upwards will decelerate at this constant rate until it comes to a stop and reverses its direction. When calculating motion, be sure to note that this acceleration is always acting downward, hence the negative sign in equations when an upward direction is considered positive. Understanding how gravity affects the motion of objects helps in predicting how long the object will be in motion before coming back to the point of origin.
Kinematic Equations
Kinematic equations are a set of formulas that describe the motion of an object under constant acceleration. They are widely used in physics to solve problems involving projectile motion like the one in this exercise.The kinematic equation used in this context is given by:
  • \( y = v_{i_y}t + \frac{1}{2} a t^2 \)
This equation relates the displacement \( y \) to the initial velocity \( v_{i_y} \), time \( t \), and acceleration \( a \). Since the ball in the exercise returns to its starting point, the net displacement \( y \) is zero.Kinematic equations are useful because they allow us to compute various unknowns in a projectile problem when some of the variables have been provided.
Initial Velocity Calculation
In problems involving projectiles, calculating the initial velocity is essential to predict the subsequent motion of the object. Specifically, the initial velocity determines how high and how far the object will travel.In our exercise, the calculation of initial velocity uses the motion equation:
  • \( 0 = v_{i_y} (4.0) + \frac{1}{2}(-1.60)(4.0)^2 \)
By substituting the values we know, such as the time of 4.0 seconds and the acceleration of \(-1.60\, \mathrm{m/s^2}\), we can rearrange the equation to find \( v_{i_y} \). Solving gives us \( v_{i_y} = 3.2 \) m/s, indicating the ball's speed just as it leaves the thrower's hand.Understanding how to manipulate these equations is fundamental for accurately determining the parameters of motion, such as initial velocity in projectile cases.

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Most popular questions from this chapter

Two balls are dropped to the ground from different heights. One is dropped \(1.5 \mathrm{~s}\) after the other, but they both strike the ground at the same time, \(5.0 \mathrm{~s}\) after the first was dropped. (a) What is the difference in the heights from which they were dropped? (b) From what height was the first ball dropped?

A body with initial velocity \(8.0 \mathrm{~m} / \mathrm{s}\) moves along a straight line with constant positive acceleration and travels \(640 \mathrm{~m}\) in \(40 \mathrm{~s}\). For the 40 s interval, find \((a)\) the average velocity, \((b)\) the final velocity, and \((c)\) the acceleration.

A stone is thrown straight downward with initial speed \(8.0 \mathrm{~m} / \mathrm{s}\) from a height of \(25 \mathrm{~m}\). Find \((a)\) the time it takes to reach the ground and \((b)\) the speed with which it strikes.

A ball is thrown upward at an angle of \(30^{\circ}\) to the horizontal and lands on the top edge of a building that is \(20 \mathrm{~m}\) away. The top edge is \(5.0 \mathrm{~m}\) above the throwing point. How fast was the ball thrown?

The speed of a train is reduced uniformly from \(15 \mathrm{~m} / \mathrm{s}\) to \(7.0 \mathrm{~m} / \mathrm{s}\) while traveling a distance of \(90 \mathrm{~m} .(a)\) Compute the acceleration. (b) How much farther will the train travel before coming to rest, provided the acceleration remains constant? Take the direction of motion to be the \(+x\) -direction. (a) We have \(v_{i x}=15 \mathrm{~m} / \mathrm{s}, v_{f x}=7.0 \mathrm{~m} / \mathrm{s}, x=90 \mathrm{~m}\). Then \(v^{2} f_{x}=v^{2} i x\) \(+2 a x\) gives \(a=-0.98 \mathrm{~m} / \mathrm{s}^{2}\) (b) The new conditions \(v_{i x}=7.0 \mathrm{~m} / \mathrm{s}, v_{f}=0\), and \(a=-0.98 \mathrm{~m} / \mathrm{s}^{2}\) now obtain. Then leads to

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