A particle is projected with a velocity v , so that its range on a horizontal plane is twice the greatest height attained. If g is the acceleration due to gravity, then its range is -

Question

A particle is projected with a velocity v , so that its range on a horizontal plane is twice the greatest height attained. If g is the acceleration due to gravity, then its range is - (A) (4 v2/5 g) (B) (4 g/5 v2) (C) (4 v2/5 g2) (D) (4 v/5 g2)

Tardigrade Admin · Accepted Answer

∵ R=2H ∴ (2 v2 s i n θ c o s θ /g)=2(v2 s i n2 θ /2 g)⇒ tanθ =2 ⇒ R=(2 v2 (s i n θ ) (. c o s θ .)/g)=(2 v2 ((2/√5) ) ((1/√5))/g)=(4 v2/5 g)

Q. A particle is projected with a velocity $v$ , so that its range on a horizontal plane is twice the greatest height attained. If $g$ is the acceleration due to gravity, then its range is -

$\frac{4 v ^{2}}{5 g}$

$\frac{4 g}{5 v ^{2}}$

$\frac{4 v ^{2}}{5 g ^{2}}$

$\frac{4 v}{5 g ^{2}}$

$∵ R = 2 H$
$∴ \frac{2 v ^{2} s in θ cos θ}{g} = 2 \frac{v ^{2} s i n ^{2} θ}{2 g} \Rightarrow t an θ = 2$
$\Rightarrow R = \frac{2 v ^{2} ( s in θ ) ( cos θ )}{g} = \frac{2 v ^{2} ( \frac{2}{5} ) ( \frac{1}{5} )}{g} = \frac{4 v ^{2}}{5 g}$

Q. A particle is projected with a velocity v , so that its range on a horizontal plane is twice the greatest height attained. If g is the acceleration due to gravity, then its range is -

5g4v2​

5v24g​

5g24v2​

5g24v​