Question

the equation of projectile is y=16x-11 the horizontal range is​

Answer 1

Answer:

Hey User

To find the horizontal range. Put value of y to be ZERO as in equation of trajectory when ∆y is ZERO x = R i.e Horizontal Range

so, 0=16x-x²/4

16x = x²/4

x = 16 × 4

x = 64m

Hope You Understood

Explanation: