91Ó°ÊÓ

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.

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 \(0.60{\rm{ s}}\).