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3-2.Motion in Plane
medium
The range of the projectile projected at an angle of $15^{\circ}$ with horizontal is $50\,m$. If the projectile is projected with same velocity at an angle of $45^{\circ}$ with horizontal, then its range will be $........\,m$
A
$50$
B
$50 \sqrt{2}$
C
$100$
D
$100 \sqrt{2}$
(JEE MAIN-2023)
Solution
$R =\frac{ v ^2 \sin 2 \theta}{ g }$
$R \propto \sin (2 \theta)$
$\frac{ R _1}{ R _2}=\frac{\sin \left(2 \theta_1\right)}{\sin \left(2 \theta_2\right)}=\frac{\sin (2 \times 15)}{\sin (2 \times 45)}=\frac{\sin 30^{\circ}}{\sin 90^{\circ}}$
$\frac{50}{ R _2}=\frac{1}{2}$
$R _2=100\,m$
Standard 11
Physics
