Question

A body is projected with a velocity u so that the horizontal range is twice the maximum height. Then the maximum height is

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

Answer 1

Answer: (D)

$\frac{2u^2}{5g}$

According to question, horizontal range is twice the maximum height

$R=2H$

$\frac{u^{2}sin2\theta }{g}=2\frac{u^{2}si{n}^2\theta }{2g}$

$2sin\theta .cos\theta =sin\theta .sin\theta$

$tan\theta =2$

Then, $sin\theta =\frac{2}{\sqrt{5}}$, $cos\theta =\frac{1}{\sqrt{5}}$

Maximum height in projectile motion,

$H=\frac{u^{2}si{n}^2\theta }{2g}$

$H=\frac{u^2}{2g}\times (\frac{2}{\sqrt{5}}{)}^2$

$H=\frac{2u^2}{5g}$

The correct option is D.