The equation of projectile is y = 16x-5x²/4. The horizontal range is

To find the horizontal range of the projectile, we start with the given equation of the trajectory:
$y=16x-\frac{5x^2}{4}$

The horizontal range ($R$) is defined as the horizontal distance traveled by the projectile when it returns to the same vertical level from which it was projected. In other words, it is the value of $x$ (other than $x=0$) for which the vertical displacement $y=0$.

Setting $y=0$ in the trajectory equation:
$16x-\frac{5x^2}{4}=0$

We can factor out $x$ from the equation:
$x\left(16-\frac{5x}{4}\right)=0$

This gives two possible solutions:
1) $x=0$ (which represents the point of projection)
2) $16-\frac{5x}{4}=0$

Solving the second equation for $x$ gives the horizontal range $R$:
$16=\frac{5x}{4}$

Multiplying both sides by 4:
$64=5x$

Dividing by 5:
$x=\frac{64}{5}=12.8\text{m}$

Therefore, the horizontal range is 12.8 m.