Question

75. Trajectory of a projectile is y=Ax-Bx? Its horizontal range
is
A
B
1) AB
2)
3)
B
4) (A2+B2
A​

Answer 1

Answer:

Equation of trajectory y= xtanθ[ 1− x/R ]

where R is the horizontal range.

So, y=ax[1− x/a/b ]

Thus, horizontal range

R= a/b

hope it was helpful : ) : ) : )

( it's y=ax−bx^2)