- Home
- Standard 11
- Physics
3-2.Motion in Plane
normal
A particle is projected with a velocity $v$ such that its range on the horizontal plane is twice the greatest height attained by it. The range of the projectile is (where $g$ is acceleration due to gravity)
A
$\frac{{4{v^2}}}{{5g}}$
B
$\frac{{4g}}{{5{v^2}}}$
C
$\frac{{{v^2}}}{g}$
D
$\frac{{4{v^2}}}{{\sqrt 5 g}}$
Solution
(a) $R = 2H$ given
We know $R = 4H\cot \theta $==> $\cot \theta = \frac{1}{2}$
From triangle we can say that $\sin \theta = \frac{2}{{\sqrt 5 }}$,$\cos \theta = \frac{1}{{\sqrt 5 }}$
$\therefore$ Range of projectile $R = \frac{{2{v^2}\sin \theta \cos \theta }}{g}$
$ = \frac{{2{v^2}}}{g} \times \frac{2}{{\sqrt 5 }} \times \frac{1}{{\sqrt 5 }}$ = $\frac{{4{v^2}}}{{5g}}$.
Standard 11
Physics
