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

\frac{a}{b}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

\frac{b}{a}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

a + b

No worries! We‘ve got your back. Try BYJU‘S free classes today!

b - a

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $\frac{a}{b}$
$y = a x - b x^{2}$
$y = a x (1 - \frac{b x}{a})$
$= a x (1 - \frac{x}{\frac{a}{b}})$
Compare the above equation from equation of trajectory of projectile
$y = x tan θ (1 - \frac{x}{R})$
We get,
$R = \frac{a}{b}$

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 = 8 x - \frac{5 x^{2}}{2}

The equation of projectile is $y = 16 x - \frac{5 x^{2}}{4}$ . The horizontal range is

Join BYJU'S Learning Program