Question

A ball is thrown with a speed $u$ , at an angle $\theta $ with the horizontal. At the highest point of its motion, the strength of gravity is somehow doubled. Taking this change into account, the total time of flight of the projectile is

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

Answer 1

Answer: (D)

Time of ascent
$t_{1}=\frac{u s i n \theta }{g}$
Time of descent
$t_{2}=\sqrt{\frac{2 H}{2 g}}$
$t_{2}=\sqrt{\frac{u^{2} s i n^{2} \theta }{2 g^{2}}}=\frac{u s i n \theta }{\sqrt{2} g}$
$t=t_{1}+t_{2}=\frac{u s i n \theta }{g}+\frac{u s i n \theta }{\sqrt{2} g}$
$\Rightarrow \frac{u s i n \theta }{g}\left(\frac{\sqrt{2} + 1}{\sqrt{2}}\right)$