Answer

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**Hint**Horizontal range is given in the question. We have to find an equation of trajectory. We will use $y = x\tan \theta \left[ {1 - \dfrac{x}{R}} \right]$ formula to calculate trajectory equation.

**Complete step by step solution:**

Parabolic trajectory:

It is Kepler’s orbit having an eccentricity equal to 1. Its orbit is unbound. For example:

When a ball is thrown upward and returns back to the ground.

Equation of motion: $y = x\tan \theta \left[ {1 - \dfrac{x}{R}} \right]$ … (1)

Horizontal range (R):

Horizontal distance covered by a particle undergoing a projectile motion.

Here x and y are axes. $\theta $ is the angle which the path of the particle makes with the horizontal.

$y = 16x - \dfrac{{5{x^2}}}{4}$ … (2)

On comparing (1) with (2), we get

$y = 16x\left[ {1 - \dfrac{x}{{\dfrac{{64}}{5}}}} \right]$

$R = \dfrac{{64}}{5} = 12.8m$

It is given in the question that horizontal range is 12.8m. So, $y = 16x - \dfrac{{5{x^2}}}{4}$ is the equation of parabolic trajectory.

**Hence part A is the correct option**

**Note**If we put the value of R in the formula as 12.8m then the equation of motion will be in terms of $\theta $ , but it is not mentioned in any of the options. If we equate equation of motion with $16x - \dfrac{{3{x^2}}}{4}$ equation then $R = \dfrac{{64}}{5} = 21.9m$ . But the given value of R is 12.8m. So, option B is wrong. If we equate equation of motion with $14x - \dfrac{{5{x^2}}}{4}$ equation then $R = \dfrac{{56}}{5} = 11.2m \ne 12.8m$ . But the given value of R is 12.8m. So, option C is wrong. If we equate equation of motion with $12x - \dfrac{{5{x^2}}}{4}$ equation then $R = \dfrac{{48}}{5} = 9.6m \ne 12.8m$ . But the given value of R is 12.8m. So, option D is wrong. Thus, we are left with option A which satisfies with the equation.

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