Question

A particle covers 50 m distance when projected with an initial speed. On the same surface it will cover a distance, when projected with double the initial speed

• A. 100 m
• B. 150 m
• C. 200 m
• D. 250 m

Answer 1

Answer: (C)

200 m

Answer 2

$R = \frac{u^{2}\sin 2\theta}{g}$ $\therefore R \propto u^{2}$ so $\frac{R_{2}}{R_{1}} = \left( \frac{u_{2}}{u_{1}} \right)^{2} = \left( \frac{2u}{u} \right)^{2}$

⇒ $R_{2} = 4R_{1}$= 4 × 50 = 200 m