Question

the equation of a projectile is y = 16x - x^2/4 the horizontal ranges?​

Answer 1

Answer:

Horizontal range is distance between between the two points where the projectile is on the ground.

Which means y = 0.

If you solve for y = 0

0 = 16x - x^2/4

x(16-x/4) = 0

x = 0 or 16-x/4 = 0

16=x/4

x = 64m

So from the given equation, the projectile is on the ground at x = 0 and x = 64m

Therefore, Range = 64m

Answer 2

Answer:

Horizontal Range(R)= 64 m

Explanation:

Given ;-

                    y = 16x - x²/4

                    y = 16x [ 1 - x²/ 4 × 16 ]

                    y = 16x [ 1 - x²/ 64 ] ______{1}

Since ;-

                    y = xtan ∅ [ 1 - x/R ] _____{2}  (from standard equation of trajectory in 2-D projectile motion.

Comparing horizontal ranges in {1} and {2}, we get;-

                   1/ R = 1/ 64

                       R = 64 m

Hence, the horizontal range ( R) = 64 m is the answer.

Hope it helps ;-))