Book edition 10th Edition

Author(s) David Halliday

Pages 1328 pages

ISBN 9781118230718

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Short Answer

At what initial speed must the basketball player in Fig. 4-50 throw the ball, at angle $θ_{0} = 55 °$ above the horizontal, to make the foul shot? The horizontal distances are $d_{1} = 1.0 ft and d_{2} = 14 ft, and the heights are h_{1} = 7.0 ft and h_{2} = 10.0 ft .$

The initial speed of the basketball is 23 ft/s

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: The given data

$a) The angle made by the basketball above the horizontal, (θ) = 55 ° b) The horizontal distances, d_{1} = 1 ft, d_{2} = 14 ft c) The heights are, h_{1} = 7 ft, h_{2} = 10 ft$

Step 2: Understanding the concept of the projectile motion

Projectile motion is the motion of a particle that is launched with an initial velocity. During its flight, the particle’s horizontal acceleration is zero and its vertical acceleration is the free-fall acceleration. (Upward is taken to be a positive direction.)

Using the equation for the projectile path in the projectile motion of the ball, we can find the initial speed of the basketball.

Formula:

The equation of the distance for the projectile path of a body,

$y = x {tanθ}_{0} - \frac{{gx}^{2}}{2 {(V_{0} {cosθ}_{0})}^{2}}$ …(i)

Step 3: Calculation of the initial speed of the basketball

The total horizontal displacement of basketball is given as:

$x = d_{2} - d_{1} = 13 ft$

The total vertical displacement of basketball is given as:

$y = h_{2} - h_{1} = 3 ft$

Rearranging the equation (i)for projectile path and substituting the above values, the velocity is given as:

$V_{0} = \frac{x}{cosθ} \sqrt{\frac{g}{2 (x tanθ - y)}} = \frac{13 ft}{\cos 55 °} \sqrt{\frac{32.15 ft / s^{2}}{2 ((13 ft) \tan 55 ° - (3 ft))}} = \frac{13 ft}{0.574} \sqrt{\frac{32 ft / s^{2}}{31.13 ft}} = 22.98 ft / s \approx 23 ft / s$

Hence, the value of the initial speed of the ball is 23ft/s.

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Fundamentals Of Physics

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Short Answer

At what initial speed must the basketball player in Fig. 4-50 throw the ball, at angle $θ_{0} = 55 °$ above the horizontal, to make the foul shot? The horizontal distances are $d_{1} = 1.0 ft and d_{2} = 14 ft, and the heights are h_{1} = 7.0 ft and h_{2} = 10.0 ft .$

Step by Step Solution

Step 1: The given data

Step 2: Understanding the concept of the projectile motion

Step 3: Calculation of the initial speed of the basketball

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Fundamentals Of Physics

Answers without the blur.

Short Answer

At what initial speed must the basketball player in Fig. 4-50 throw the ball, at angleθ0=55°above the horizontal, to make the foul shot? The horizontal distances are d1=1.0ft and d2=14 ft, and the heights are h1=7.0 ft and h2=10.0ft.

Step by Step Solution

Step 1: The given data

Step 2: Understanding the concept of the projectile motion

Step 3: Calculation of the initial speed of the basketball

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At what initial speed must the basketball player in Fig. 4-50 throw the ball, at angle $θ_{0} = 55 °$ above the horizontal, to make the foul shot? The horizontal distances are $d_{1} = 1.0 ft and d_{2} = 14 ft, and the heights are h_{1} = 7.0 ft and h_{2} = 10.0 ft .$