Question

The equation of trajectory of a projectile is \[y=\sqrt{3}x-\frac{g}{8}x^{2}\] where x, y are in metre and g is in \[\frac{m}{s^{2}}.\] The initial velocity of projectile 'u' is ______ \[\left(\tan60^{\circ}=\sqrt{3},\sin60^{\circ}=\frac{\sqrt{3}}{2},\cos60^{\circ}=\frac{1}{2}\right)\] (g = acceleration due to gravity)

पर्याय

• 2 m/s

• 4 m/s

• 8 m/s

• 16 m/s

Answer 1

The equation of trajectory of a projectile is \[y=\sqrt{3}x-\frac{g}{8}x^{2}\] where x, y are in metre and g is in \[\frac{m}{s^{2}}.\] The initial velocity of projectile 'u' is 4 m/s \[\left(\tan60^{\circ}=\sqrt{3},\sin60^{\circ}=\frac{\sqrt{3}}{2},\cos60^{\circ}=\frac{1}{2}\right)\] (g = acceleration due to gravity)

Explanation:

Equation of projectile is given as

\[\mathrm{y=\sqrt{3}~x-\frac{g}{8}x^{2}}\]

Comparing with standard equation of a projectile,

\[\mathrm{y}=\mathrm{x}\tan\theta-\frac{\mathrm{g}}{2\mathrm{u}^2\cos^2\theta}\mathrm{x}^2\]

\[\tan\theta=\sqrt{3}\Rightarrow\theta=60^{\circ}\]

and \[2\mathrm{u}^2\mathrm{cos}^2\theta=8\]

\[2u^2cos^260=8\]

u = 4 m/s