Question

A body is projected with velocity $u$ such that its horizontal range and maximum vertical heights are same. The maximum heights is:

• A. $\frac{u^2}{2g}$
• B. $\frac{3u^2}{4g}$
• C. $\frac{16u^2}{17g}$
• D. $\frac{8u^2}{17g}$

Answer 1

The correct option is D $\frac{8u^2}{17g}$

Lets consider $u=$ initial speed of the projectile

$\theta =$ angle of projection

According to question,

$H=R$

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

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

$sin\theta =4cos\theta$

$tan\theta =4$

As we know that, $1+ta{n}^2\theta =se{c}^2\theta$

$\frac{1}{co{s}^2\theta }=1+16=17$

$co{s}^2\theta =\frac{1}{17}$

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

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

$si{n}^2\theta =1-\frac{1}{17}=\frac{16}{17}$

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

$H=\frac{u^2}{2g}\frac{16}{17}$

$H=\frac{8u^2}{17g}$

The correct option is D.