Question

A particle is projected from ground at some angle to the horizontal. $($Assuming point of projection to be origin and the horizontal and vertical directions to be the $x\&y$ axis$)$ the particle passes through the points $(3,4)$ and $(4,3)$ during its motion. If the range of the particle is $\frac{37}{x}m$ then $x$ is:

Answer 1

The equation of trajectory:

$y=x\tan \alpha (1-\frac{x}{R})$

Substituting the point (3,4)
$4=3\tan \alpha (1-\frac{3}{R})-------\left(i\right)$

Substituting the point (4,3)
$3=4\tan \alpha (1-\frac{4}{R})-------\left(ii\right)$

Dividing $\left(i\right)$ and $\left(ii\right)$, we get :

$\frac{4}{3}=\frac{3}{4}\frac{1-\frac{3}{R}}{1-\frac{4}{R}}$

$\Rightarrow 16(1-\frac{4}{R})=9(1-\frac{3}{R})$

$\Rightarrow 7=\frac{64}{R}-\frac{27}{R}$

$\Rightarrow R=\frac{37}{7}$