Question

For an object projected from the ground with speed $u$. The horizontal range is two times the maximum height attained by it. The horizontal range of the object is:

• A. $\frac{2u^2}{3g}$
• B. $\frac{3u^2}{4g}$
• C. $\frac{3u^2}{2g}$
• D. $\frac{4u^2}{5g}$

Answer 1

The correct option is D $\frac{4u^2}{5g}$

In projectile motion,

The maximum height is $H=\frac{u^{2}si{n}^2\theta }{2g}$

Horizontal range, $R=\frac{u^{2}sin2\theta }{g}$

According to question,

$R=2H$

$\frac{2u^{2}sin\theta .cos\theta }{g}=2\frac{u^{2}si{n}^2\theta }{2g}$

$2cos\theta =sin\theta$

$tan\theta =2$

We know that, $1+ta{n}^2\theta =se{c}^2\theta =\frac{1}{co{s}^2\theta }$

$1+2^2=\frac{1}{co{s}^2\theta }$

$co{s}^2\theta =\frac{1}{5}$

$cos\theta =\frac{1}{\sqrt{5}}$

And, $si{n}^2\theta =1-co{s}^2\theta$

$si{n}^2\theta =1-\frac{1}{5}=\frac{4}{5}$

$sin\theta =\frac{2}{\sqrt{5}}$

Horizontal range,

$R=\frac{2u^{2}sin\theta .cos\theta }{g}$

$R=\frac{2u^2}{g}\frac{2}{\sqrt{5}}\frac{1}{\sqrt{5}}$

$R=\frac{4u^2}{5g}$

The correct option is D.