Question

The maximum horizontal distance covered by the ball will be? If angle of projection is ${30}^{\circ }$.

• A. $160\sqrt{3}m$
• B. $140\sqrt{3}m$
• C. $120\sqrt{3}m$
• D. $100\sqrt{3}m$

Answer 1

Answer: (A)

$160\sqrt{3}m$

$R=\frac{u^2\sin 2\theta }{g}=\frac{(56{)}^2\sin {60}^{\circ }}{9.8}$
$=\frac{56\times 56\times \sqrt{3}}{19.6}=160\sqrt{3}m$.