The equation of motion of a projectile is y = | vedclass

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Question

$\because \quad \mathrm{y}=\mathrm{Ax}-\mathrm{Bx}^{2}$

at $\quad x=R, y=0$

and $x=0, y=0$

so put $y=0$

${\mathrm{Ax}-\mathrm{Bx}^{2}=0}$

Vedclass · Accepted Answer

$\frac{A}{B}$

Vedclass · Answer

$\because \quad \mathrm{y}=\mathrm{Ax}-\mathrm{Bx}^{2}$

at $\quad x=R, y=0$

and $x=0, y=0$

so put $y=0$

${\mathrm{Ax}-\mathrm{Bx}^{2}=0}$

${\mathrm{x}(\mathrm{A}-\mathrm{Bx})=0}$

$x=0$ or $A-B x=0$

for $x=0,$ initial point

and $A-B x=0 \rightarrow x=\frac{A B}{B}$ or $R=\frac{A}{B}$

The equation of motion of a projectile is $y = Ax -Bx^2$ where $A$ and $B$ are the constants of motion. The horizontal range of the projectile is

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