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Q3.4-30 PE

Expert-verifiedFound in: Page 87

Book edition
1st Edition

Author(s)
Paul Peter Urone

Pages
1272 pages

ISBN
9781938168000

**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.

(c) The time of flight is \(0.60\;{\rm{s}}\).

**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.**

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

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 \(14.2^\circ \).

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.

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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