Question

y=16x-5x^2/4 find the horizontal range?

Answer 1

Do you know that x=0 and x= R, y=0? Do you get this, why is it so? Because the projectile is at the surface of projection at x=0 while at x=R also the projectile​ comes to the surface of projection after flight, hence y=0.

we just have to find the roots of y= 16x-5x²/4= 0

16x -5x²/4 = 0

x(16 -5x/4) = 0

✓ x=0 and

✓ 16-5x/4=0

ie x = 64/5 = 12.8 units

Answer 2

Heya ____

y=16x-5x^2/4 find the horizontal range???

• We know that x = 0 , x = r then y =>0 due to the projectile on the surface ______

• Find the roots of y ____

16x - 5x ^2 /4 = 0

x(16-5x/4) = 0

16-5x/4 = 0

x = 64/5

= 12.8 ___ Ans

Thank you