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. $\frac{3}{4}\frac{u^{2}sin2\theta }{g}$
• B. $\frac{u^{2}sin2\theta }{2g}\left(1+\frac{1}{\sqrt{2}}\right)$
• C. $\frac{u^2}{g}sin2\theta$
• D. $\frac{u^{2}sin2\theta }{2g}\left(2+\frac{1}{\sqrt{2}}\right)$

Answer 1

Answer: (B)

The time taken to reach maximum height and maximum height are $t=\frac{usin\theta }{g}$ and $H=\frac{u^{2}si{n}^2\theta }{2g}$
For remaining half, the time of flight is
$t^{\prime }=\sqrt{\frac{2H}{\left(2g\right)}}$
$=\sqrt{\frac{u^{2}si{n}^2\theta }{2g^2}}$
$=\frac{t}{\sqrt{2}}$
$\therefore$ Total time of flight is $t+t^{\prime }$,
$T=t\left(1+\frac{1}{\sqrt{2}}\right)$
$T=\frac{usin\theta }{g}\left(1+\frac{1}{\sqrt{2}}\right)$
So horizontal range is $ucos\theta \times T$
$=\frac{u^{2}si{n}^2\theta }{2g}\left(1+\frac{1}{\sqrt{2}}\right)$