A particle is projected with velocity $2 \sqrt{g h}$, so that it just flies over two walls of equal height…
A particle is projected with velocity $2 \sqrt{g h}$, so that it just flies over two walls of equal height $h$ and $2 h$ distance apart from each other. Find the time for which the particle flies between the walls.
$\sqrt{\frac{4 h}{g}}$
$\sqrt{\frac{h}{g}}$
$\sqrt{\frac{4 g}{h}}$
$\sqrt{\frac{g}{h}}$
Solution
Velocity of projection,
$
\begin{aligned}
v & =2 \sqrt{g h} \\
\therefore \quad v_x & =v \cos \theta=2 \sqrt{g h} \cos \theta
\end{aligned}
$
$\therefore$ Time taken by the projectile to cover interwall distance,
$
\begin{aligned}
t & =\frac{2 h}{v_x}=\frac{2 h}{2 \sqrt{g h} \cos \theta} \\
\Rightarrow \quad t & =\sqrt{\frac{h}{g}} \cdot \sec \theta
\end{aligned}
$
Vertical velocity at the top of the wall is given as
$
\begin{aligned}
v_y^{\prime 2} & =v_y^2-2 g h=(\sqrt{2 g h} \sin \theta)^2-2 g h \\
& =4 g h \sin ^2 \theta-2 g h \\
v_y^{\prime 2} & =2 g h\left(2 \sin ^2 \theta-1\right) \\
\therefore \quad v_y^{\prime} & =\sqrt{2 g h\left(2 \sin ^2 \theta-1\right)} \\
& t=\frac{2 v_y^{\prime}}{g} \\
t & =\frac{2 \sqrt{2 g h\left(2 \sin ^2 \theta-1\right)}}{g}
\end{aligned}
$
From Eqs. (i) and (ii), we get
$
\sqrt{\frac{h}{g}} \sec \theta=\frac{2 \sqrt{2 g h\left(2 \sin ^2 \theta-1\right)}}{g}
$
Squaring both side,
$
\begin{aligned}
& \Rightarrow \frac{h}{g} \sec ^2 \theta=\frac{4 \times 2 g h\left(2 \sin ^2 \theta-1\right)}{g^2} \\
& \Rightarrow 1=8 \cos ^2 \theta\left(2 \sin ^2 \theta-1\right) \\
& \Rightarrow 1=8 \cos ^2 \theta\left[2\left(1-\cos ^2 \theta\right)-1\right]
\end{aligned}
$
$
\begin{array}{cc}
\Rightarrow & 16 \cos ^4 \theta-8 \cos ^2 \theta+1=0 \\
& \left(4 \cos ^2 \theta-1\right)^2=0 \\
\Rightarrow & \cos ^2 \theta=\frac{1}{4} \\
\Rightarrow & \cos \theta=\frac{1}{2}=\cos 60^{\circ} \Rightarrow \theta=60^{\circ}
\end{array}
$
$\therefore$ From Eq. (i),
$
t=\sqrt{\frac{h}{g}} \sec 60^{\circ}=\sqrt{\frac{h}{g}} \cdot 2=\sqrt{\frac{4 h}{g}}
$