JEE Main · 2023 · 10 Apr 2023 · Shift 1 · Kinematics

The range of the projectile projected at an angle of

1 5^{\circ}

with horizontal is 50 m. If the projectile is projected with same velocity at an angle of

4 5^{\circ}

with horizontal, then its range will be:

(A)50 m
(B) $502 m$
(C)100 m✓ Correct
(D) $1002 m$

Answer: (C)

Solution

The range is

R = \frac{v ^{2} sin 2 θ}{g}

, so for a fixed launch speed

R \propto sin 2 θ

. Thus

\frac{R _{15}}{R _{45}} = \frac{sin 3 0 ^{\circ}}{sin 9 0 ^{\circ}} = \frac{1/2}{1}

. With

R_{15} = 50

m, the new range is

R_{45} = 2 \times 50 = 100

JEE Main · 2023 · 10 Apr 2023 · Shift 1 · Single correct option

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