Book edition 1st Edition

Author(s) Paul Peter Urone

Pages 1272 pages

ISBN 9781938168000

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Short Answer

A rugby player passes the ball \(7.00{\rm{ m}}\) across the field, where it is caught at the same height as it left his hand.
(a) At what angle was the ball thrown if its initial speed was \(12.0{\rm{ m}}/{\rm{s}}\), assuming that the smaller of the two possible angles was used?
(b) What other angle gives the same range, and why would it not be used?
(c) How long did this pass take?

(a) The smaller angle is \(14.2^\circ \).

(b) The other angle is \(75.8^\circ \), and it is not used because a high angle increases the chance of interception.

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Projectile motion

When a projectile is launched at an angle to the horizontal, a continuous acceleration acts in the vertical direction on the projectile, which is pointed downwards, and there is no horizontal acceleration. Projectile motion is the name for this type of motion.

Step 2: Given data

Range of the projectile \(R = 7.00\,{\rm{m}}\).
Initial speed \(u = 12.0\,{\rm{m}}/{\rm{s}}\).

Step 3: (a) Angle of projection

The horizontal range of the projectile motion is,

\(R = \frac{{{u^2}\sin \left( {2\theta } \right)}}{g}\)

Here \(R\) is the horizontal range of the projectile motion, \(u\) is the initial velocity of the projectile, \(\theta \) is the angle of projection and \(g\) is the angle of projection.

Rearranging the above equation in order to get an expression for the angle of projection,

\(\theta = \frac{1}{2}{\sin ^{ - 1}}\left( {\frac{{Rg}}{{{u^2}}}} \right)\)

Substitute \(7.00{\rm{ m}}\) for \(R\), \(9.8{\rm{ m}}/{{\rm{s}}^2}\) for \(g\), and \(12.0{\rm{ m}}/{\rm{s}}\) for \(u\),

\(\begin{aligned}{}\theta = \frac{1}{2}{\sin ^{ - 1}}\left( {\frac{{\left( {7.00{\rm{ m}}} \right) \times \left( {9.8{\rm{ m}}/{{\rm{s}}^2}} \right)}}{{{{\left( {12.0{\rm{ m}}/{\rm{s}}} \right)}^2}}}} \right)\\ = 14.2^\circ \end{aligned}\)

Hence, a smaller angle of projection is \(14.2^\circ \).

Step 4: (b) Another angle for the same range

We know that,

\(\sin \theta = \sin \left( {90^\circ + \varphi } \right)\)

Therefore,

\(\begin{aligned}{}\theta = 90^\circ + \varphi \\\varphi = 90^\circ - \theta \end{aligned}\)

Substitute \(14.2^\circ \) for \(\theta \),

\(\begin{aligned}{}\varphi = 90^\circ - 14.2^\circ \\ = 75.8^\circ \end{aligned}\)

Hence, the other angle giving the same range is \(75.8^\circ \), and it is not used because a high angle increases the chance of interception.

Step 5: (c) Time of projectile

The time of the projectile is,

\(t = \frac{{2u\sin \theta }}{g}\)

Substitute \(12.0{\rm{ m}}/{\rm{s}}\) for \(u\), \(14.2^\circ \) for \(\theta \), and \(9.8{\rm{ m}}/{{\rm{s}}^2}\) for \(g\),

\(\begin{aligned}{}t = \frac{{2 \times \left( {12.0{\rm{ m}}/{\rm{s}}} \right) \times \sin \left( {14.2^\circ } \right)}}{{\left( {9.8{\rm{ m}}/{{\rm{s}}^2}} \right)}}\\ = 0.60{\rm{ s}}\end{aligned}\)

Hence, the time of flight is \(0.60{\rm{ s}}\).

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College Physics (Urone)

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Short Answer

Step by Step Solution

Step 1: Projectile motion

Step 2: Given data

Step 3: (a) Angle of projection

Step 4: (b) Another angle for the same range

Step 5: (c) Time of projectile

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College Physics (Urone)

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Short Answer

Step by Step Solution

Step 1: Projectile motion

Step 2: Given data

Step 3: (a) Angle of projection

Step 4: (b) Another angle for the same range

Step 5: (c) Time of projectile

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