A ball is projected with a velocity $5 \mathrm{~m} / \mathrm{s}$, so that its horizontal range is twice the greatest height attained. The value of range is

• A

  10 m
• Correct answer
  B

  2 m
• C

  4 m
• D

  5 m

Solution

Given that, initial velocity, $u=5 \mathrm{~m} / \mathrm{s}$ and horizontal range $=2 \times$ greatest height attained or $\begin{array}{ll} \text { or } & H=2 \times R \Rightarrow \frac{u^2 \sin ^2 \theta}{2 g}=\frac{u^2 \sin 2 \theta}{g} \\ \Rightarrow & \frac{\sin ^2 \theta}{2}=\sin 2 \theta \Rightarrow \frac{\sin ^2 \theta}{2}=2 \sin \theta \cos \theta \\ \Rightarrow & \frac{\sin \theta}{2}=2 \cos \theta \Rightarrow \frac{\sin \theta}{\cos \theta}=4 \end{array}$ $\Rightarrow \quad \tan \theta=4 \Rightarrow \tan \theta=\frac{4}{1}=\frac{\text { Perpendicular }}{\text { Base }}$ If we form a right angled triangle, with these sides, then
Now, $\sin \theta=\frac{4}{\sqrt{17}}$ and $\cos \theta=\frac{1}{\sqrt{17}}$ So, the horizontal range $\begin{aligned} R & =\frac{u^2 \sin 2 \theta}{g}=\frac{u^2 \times 2 \sin \theta \cos \theta}{g} \\ & =\frac{(5)^2 \times 2 \times \frac{4}{\sqrt{17}} \times \frac{1}{\sqrt{17}}}{10}=\frac{25 \times 2 \times 4}{10 \times 17}=\frac{20}{17} \\ & =1.17 \mathrm{~m} \approx 2 \mathrm{~m} \end{aligned}$

Hence, the correct answer is option: 'B'.