Question

A tank is filled up to height $2H$ with a liquid and is placed on a platform of height $H$ from the ground. The distance $x$ from the ground is ${}^{\prime }{Z}^{\prime }H$ where a small hole is punched to get the maximum range $R$ then the value of ${}^{\prime }{Z}^{\prime }$ is

Answer 1

As we know, Range $R=v\times t$ (where $v$ is velocity and $t$ is time)
If we think in term of projectile motion then we can say
$x=\frac{1}{2}g{t}^2$
$t=\sqrt{\frac{2x}{g}}$
The velocity of flux can be writen as$v=\sqrt{2g(3H-x)}$ where, $(3H-x)$ is the height of the water lable above hole
Now, $R=\sqrt{2g(3H-x)}\times \sqrt{\frac{2x}{g}}$
$R=\sqrt{12Hx-4x^2}$ Here, $R=f\left(x\right)$
From above equation we can visualise that this is a quadratic equation and if we plot graph between $R$ and $x$
Due to symmetry, the maximum value of range will be at the centre and that will be at $x=\frac{3H}{2}$
$⟹z=\frac{3}{2}=1.5$