The equation of a projectile is $y = a x - b x^{2}$ . Its horizontal range is

Open in App

Solution

Equation of trajectory $y = x tan θ [1 - \frac{x}{R}]$
where $R$ is the horizontal range.
So, $y = a x [1 - \frac{x}{a / b}]$
Thus, horizontal range $R = \frac{a}{b}$

Suggest Corrections

Similar questions

y = a x - b x^{2}

y = A x - B x^{2}

The equation of projectile is y=ax-bx^2 then horizontal range is

y = 16 x - \frac{x^{2}}{4}

y = 16 x - \frac{5 x^{2}}{4}

Join BYJU'S Learning Program