Question

The equation of projectile is y = $\sqrt{3}x$– $\frac{g}{2}x^{2}$. The angle of projection and initial velocity is –

• A. 30^o, 4 m/s
• B. 60^o, 2 m/s
• C. 60^o, 4 m/s
• D. 90^o, 4 m/s

Answer 1

Answer: (B)

60^o, 2 m/s

Answer 2

Standard equation of projectile is

y = $(\tan\theta)x–\frac{g}{2u^{2}\cos^{2}\theta}x^{2}$ …..(1)

given equation is

y = $\sqrt{3}x–\frac{g}{2}x^{2}$ …..(2)

Comparing (1) and (2)

tan q = $\sqrt{3}$ ̃ q = 60^o

and 2u²cos²q = 2

u = $\frac{1}{\cos\theta} = \frac{1}{\cos 60{^\circ}}$= 2 m/s