A ball is thrown with a speed u , at an angle θ 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

Question

A ball is thrown with a speed u , at an angle θ 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) (2 u s i n θ /g) (B) (3 u s i n θ /2 g) (C) (3 u sin θ /4 g) (D) ((√2 + 1/√2))(u sin θ /g)

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="Solution" src="https://cdn.tardigrade.in/q/nta/p-uyj6bokbedjl.png" /> Time of ascent t1=(u s i n θ /g) Time of descent t2=√(2 H/2 g) t2=√(u2 s i n2 θ /2 g2)=(u s i n θ /√2 g) t=t1+t2=(u s i n θ /g)+(u s i n θ /√2 g) ⇒ (u s i n θ /g)((√2 + 1/√2))

Q. A ball is thrown with a speed $u$ , at an angle $θ$ 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

$\frac{2 u s in θ}{g}$

$\frac{3 u s in θ}{2 g}$

$\frac{3 u s in θ}{4 g}$

$(\frac{2 + 1}{2}) \frac{u s in θ}{g}$

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

Q. A ball is thrown with a speed u , at an angle θ 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

g2usinθ​

2g3usinθ​

4g3usinθ​

(2​2​+1​)gusinθ​