Question

A body is projected with velocity '$u$' so that the maximum height is thrice the horizontal range. Then the maximum height is

• A. $\frac{72u^2}{145g}$
• B. $\frac{72u^2}{\sqrt{145}g}$
• C. $\frac{u^2}{2g}$
• D. $\frac{145u^2}{72g}$

Answer 1

The correct option is B $\frac{72u^2}{\sqrt{145}g}$

Initial velocity$=4$

$H_{max}=3$ Range

To find: Maximum Height$=?$

Solution:-

$\frac{u^2{\sin }^2\theta }{2g}=\frac{3u^2\sin 2\theta }{g}$

$\frac{{\sin }^2\theta }{2}=3\times 2\sin \theta \cos \theta$

$\tan \theta =12$

$\sin \theta =\frac{12}{\sqrt{145}}$

$H_{max}=\frac{u^2{\sin }^2\theta }{2g}$

$=\frac{u^2}{2g}{\left(\frac{12}{\sqrt{145}}\right)}^2$

$H_{max}=\frac{72u^2}{\sqrt{145}g}$.