# Equation of trajectory for a projectile is y = 2x{1-(x/40)} (where x and y are in meter). What will be the range of the projectile?

Nov 1, 2016

Let the velocity of projection of the projectile in x-y plane from the origin (0,0) be u with angle of projection $\alpha$ with the horizontal direction (x-axis).
The vertical component of the velocity (along y-axis) of projection is $u \sin \alpha$ and the horizontal component is $u \cos \alpha$

Now if the time of flight be T then the object will return to the ground after T sec and during this T sec its total vertical displacement h will be zero. So applying the equation of motion under gravity we can write
$h = u \sin \alpha \times T + \frac{1}{2} g {T}^{2}$
$\implies 0 = u \times T - \frac{1}{2} \times g \times {T}^{2}$
where $g = \text{acceleration due to gravity}$
$\therefore T = \frac{2 u \sin \alpha}{g}$
The horizontal displacement during this T sec or the Range ,$R = u \cos \alpha \times T$

$\implies R = \frac{2 {u}^{2} \sin \alpha \cos \alpha}{g} \ldots \left(1\right)$

Let the position of the projectile in x-y plane after t sec of its projection be $\left(x , y\right)$

The horizontal displacement during t sec

$x = u \cos \alpha \times t \ldots . \left(2\right)$

And the vertical displacement during t sec

$y = u \sin \alpha \times t - \frac{1}{2} \times g \times {t}^{2.} \ldots . \left(3\right)$

Combining (1) and (2) we get

$y = u \sin \alpha \times \frac{x}{u \cos \alpha} - \frac{g {x}^{2}}{2 {u}^{2} {\cos}^{2} \alpha}$

$\implies y = x \tan \alpha - {x}^{2} / \left(\frac{2 {u}^{2} {\cos}^{2} \alpha}{g}\right) \ldots . \left(4\right)$

This is the equation of the trajectory,we obtained.

Now in the given problem the equation of the trajectory is

$y = 2 x \left(1 - \frac{x}{40}\right)$

$\implies y = 2 x - {x}^{2} / 20. \ldots . . \left(5\right)$

Comparing (4) and (5) we get

$\tan \alpha = 2. \ldots . \left(6\right)$

and

$\frac{2 {u}^{2} {\cos}^{2} \alpha}{g} = 20. \ldots . \left(7\right)$

Multiplying (6) by (7) we get

$\frac{\tan \alpha \times 2 {u}^{2} {\cos}^{2} \alpha}{g} = 2 \times 20$

$\implies \frac{2 {u}^{2} \sin \alpha \cos \alpha}{g} = 40$

$\implies R = 40$

So Range of the projectile $R = 40 m$