Question

A stone is projected from level ground with speed $u$ and at an angle $\theta$ with horizontal. Some how the acceleration due to gravity $(g)$ becomes double (that is $2g$) immediately after the stone reaches the maximum height and remains same thereafter. Assuming direction of acceleration due to gravity to be vertically downwards. The horizontal range of the particle is

• A. $\frac{3}{4} \frac{u^{2}\,sin\,2\theta}{g}$
• B. $ \frac{u^{2}\,sin\,2\theta}{2g}\left(1+\frac{1}{\sqrt{2}}\right)$
• C. $ \frac{u^{2}}{g}sin\,2\theta $
• D. $ \frac{u^{2}\,sin\,2\theta}{2g}\left(2+\frac{1}{\sqrt{2}}\right)$

Answer 1

Answer: (B)

The time taken to reach maximum height and maximum height are $t= \frac{u\,sin\,\theta}{g}$ and $H=\frac{u^{2}\,sin^{2}\,\theta }{2g}$
For remaining half, the time of flight is
$t'=\sqrt{\frac{2H}{\left(2g\right)}}$
$=\sqrt{\frac{u^{2}\,sin^{2}\,\theta }{2g^{2}}}$
$=\frac{t}{\sqrt{2}}$
$\therefore $ Total time of flight is $t+t'$,
$T=t\left(1+\frac{1}{\sqrt{2}}\right)$
$T=\frac{u\,sin\,\theta}{g}\left(1+\frac{1}{\sqrt{2}}\right)$
So horizontal range is $u\,cos \theta \times T$
$=\frac{u^{2}\,sin^{2}\,\theta }{2g}\left(1+\frac{1}{\sqrt{2}}\right)$