Question

the equation of projectile is y= 16x-x²/4the horizontal range is--------+ (using formula u2×sin2theta g​

Answer 1

Step-by-step explanation:

Horizontal range is when the parabola cross the axis. it is going to cut x -axis at two points and the distance between them is the range.

$y = 16x - \frac{ {x}^{2} }{4} \\ put \: y \: = 0 \\ 0 = 16x - \frac{ {x}^{2} }{4} \\ x(16 - \frac{x}{4} ) = 0 \\ = > x = 0 \: or \: 64$

x = 0 and 64 are two roots of above equation

therefore range is R = 64 − 0 = 64m

Answer 2

Answer:

Solution :

y=16x−x2y

y=16x(1−x64)

Comparing with y=xtanθ(1−xR)

R=64m

Step-by-step explanation:

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.

have a nice day