Question

a particle is projected in vertical plane follows the path 3y=4x-15x^2. the horizontal range of the particle is(g=10).

Answer 1

In this question we use equation of trajectory an compare the term

Answer 2

we know, if any particle is projected with speed u m/s at an angle $\theta$ with horizontal then, particle follows projectile trajectory. e.g., $y=xtan\theta-\frac{gx^2}{2u^2cos^2\theta}$

in question, A particle is projected in vertical plane follows the path 3y = 4x - 15x²
y = 4x/3 - 5x²
but we have standard formula in horizontal plane. so, just take $(90-\theta)$ in place of $\theta$ to convert in vertical plane of standard equation.
e.g., $y=tan(90-\theta)x-\frac{gx^2}{2u^2cos^2(90-\theta)}$

$y=cot\theta x-\frac{gx^2}{2u^2sin^2\theta}$
now compare the equations ,
$cot\theta=\frac{4}{3}$
so, $sin\theta=\frac{3}{5}$

$-5=-\frac{g}{2u^2sin^2\theta}$
5 = 10/{2u²(3/5)²}
u = 5/3 m/s

so, range = $\frac{u^2sin2\theta}{g}$
= $\frac{(5/3)^2\times2sin\theta.cos\theta}{g}$
= $\frac{(5/3)^2\times2\times3/5\times4/5}{10}$

= 24/90 m