Book edition 1st Edition

Author(s) Paul Peter Urone

Pages 1272 pages

ISBN 9781938168000

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Short Answer

A football player punts the ball at a $45.0$ angle. Without an effect from the wind, the ball would travel $60.0 m$ horizontally. (a) What is the initial speed of the ball? (b) When the ball is near its maximum height it experiences a brief gust of wind that reduces its horizontal velocity by $1.50 m / s$ . What distance does the ball travel horizontally?

(a) Initial speed of the ball is $24.2 m / s .$

(b) The total displacement will be $57.2$ meter.

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Definition of initial velocity

When gravity first exerts force on an item, its initial velocity indicates how fast it travels. The final velocity, on the other hand, is a vector number that measures a moving body's speed and direction after it has reached its maximum acceleration.

Step 2: given data

The range of the football is 60 m.

The angle of the football is released at the angle 45 degree.

The gravitational acceleration is -9.8 m/s²

The initial velocity we have to calculate.

Step 3: The initial velocity of the ball

The velocity in the x frame will be constant in all time. Considering the initial velocity

$\begin{array}{rcl} R_{x} & = & \frac{{V_{i}}^{2} s i n 2 θ_{i}}{g} \\ 60 & = & \frac{{V_{i}}^{2} s i n 2 (45)}{9.8} \\ V_{i} & = & \sqrt{588} \\ V_{i} & = & 24.2 m / s \end{array}$

The initial velocity will be positive, that is 24.2 m/s.

Step 4: Distance travelled by the ball

The Components of the initial velocity.

$\begin{array}{rcl} V_{i x} & = & V_{i} c o s θ \\ V_{i x} & = & (24.2) c o s 45 \\ V_{i x} & = & 17.1 m / s \end{array}$

The initial velocity in the x frame for part a is $17.1 m / s$ .

The velocity in the direction for frame B will be

$\begin{array}{rcl} V_{x b} & = & 17.1 - 1.50 \\ = & 15.6 m / s \end{array}$

Time is needed for finding the distance

The Y- Components of the initial velocity.

$\begin{array}{rcl} V_{i y} & = & V_{i} s i n θ \\ V_{i y} & = & (24.2) s i n 45 \\ V_{i y} & = & 17.1 m / s \end{array}$

The initial velocity in the y frame for part a is - $17.1 m / s$ . As the ball is decreasing so value of velocity became negative.

Putting the value of the given data in the equation

$\begin{array}{rcl} V_{f} & = & V_{i} + a t \\ - 17.1 & = & 17.1 + (- 9.8) t \\ t & = & 3.49 s \end{array}$

The time is total time 3.49 s, but for going from O to A and then from A to B the time becomes half $t_{a} = \frac{3.49}{2} = 1.75 s$

The displacement for part A is

$\begin{array}{rcl} \bar{V_{a}} & = & \frac{X_{a}}{t_{a}} \\ X_{a} & = & 17.1 \times 1.75 \\ X_{a} & = & 29.9 m \end{array}$

The displacement for part A is $29.9$ meter

The displacement for part B is

$\begin{array}{rcl} \bar{V_{b}} & = & \frac{X_{b}}{t_{b}} \\ X_{b} & = & 15.6 \times 1.75 \\ X_{b} & = & 27.3 m \end{array}$

The displacement for part is $27.3$ meter.

The total displacement will be $29.9 + 27.3 = 57.2$ meter.

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College Physics (Urone)

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Short Answer

Step by Step Solution

Step 1: Definition of initial velocity

Step 2: given data

Step 3: The initial velocity of the ball

Step 4: Distance travelled by the ball

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College Physics (Urone)

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Short Answer

Step by Step Solution

Step 1: Definition of initial velocity

Step 2: given data

Step 3: The initial velocity of the ball

Step 4: Distance travelled by the ball

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