9. \(
RTan\theta = 4{\rm H} = \frac{1}
{2}gT^2
\)
range of a projectile is maximum when angle of projection 45° ,
\(
\therefore R = \frac{{u^2 \sin 2\theta }}
{g}
\),R is maximum if sin2\(\theta\)
is maximm i.e., If 2\(\theta\)= 90° or \(\theta\)= 45°
range of a projectile = maximum height if \(
\theta = \tan ^{ - 1}
\) (4 )or 76°
Sol: we know that tan\(\theta\) = \(
\frac{{4H}}
{R}
\) from this
R = H\(\Rightarrow\) tan\(\theta\) = 4 \(
\Rightarrow \tan ^{ - 1} 4
\) = 76°
10.For projectile, in the case of complimentary angles,
a) Ranges are same
Sol. If \(\theta\) and (90-\(\theta\)) are angles of projection, we have
\(R_1 = \frac{{u^2 \sin 2\theta }}
{g}
\) and \(
R_2 = \frac{{u^2 \sin 2(90 - \theta )}}
{g}
\)
\(
\Rightarrow R_1 = \frac{{u^2 \sin 2\theta }}
{g}
\) and \(
R_2 = \frac{{u^2 \sin (180 - 2\theta )}}
{g}
\)
\(
\Rightarrow R_1 = R_2
\)
b).If T1,T2 are times of flight, T1T2=\(
\frac{{2R}}
{g}
\)
Sol: We have
\(
T_1 = \frac{{2u\sin \theta }}
{g}
\) and \(
T_2 = \frac{{2u\sin (90 - \theta )}}
{g}
\)
\(
\Rightarrow T_1 T_2 = \frac{{4u^2 \sin \theta \cos \theta }}
{{g^2 }} = \frac{{2u^2 \sin 2\theta }}
{{g^2 }}
\)
\(
\Rightarrow T_1 T_2 = \frac{{2R}}
{g}
\)
\(
\therefore R = \frac{1}
{2}gT_1 T_2
\)
11. \(
\begin{gathered}
y = \left( {\tan \theta } \right)x - \frac{g}
{{2u^2 \cos ^2 \theta }}x^2 \hfill \\
y = x\tan \theta \left( {1 - \frac{g}
{{2u^2 \cos ^2 \theta \tan \theta }}x} \right) \hfill \\
= x\tan \theta \left( {1 - \frac{g}
{{2u^2 \cos \theta \sin \theta }}x} \right) \hfill \\
y = x\tan \theta \left( {1 - \frac{x}
{R}} \right) \hfill \\
\end{gathered}
\)
Note :
If horizontal and vertical displacements of a projectile are respectively x = at and \(
y = bt - ct^2
\), then velocity of projection u = \(
\sqrt {a^2 + b^2 }
\) and angle of projection \(
\theta = \tan ^{ - 1} \frac{b}
{a}
\)