A projectile is projected at $30^{\circ}$ from horizontal with initial velocity $40 \mathrm{~ms}^{-1}$. The velocity of the projectile at $\mathrm{t}=2 \mathrm{~s}$ from the start will be: (Given $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}$)

(A) $20 \sqrt{3} \mathrm{~ms}^{-1}$

(B) $40 \sqrt{3} \mathrm{~ms}^{-1}$

JEE Main · 2023 · 11 Apr 2023 · Shift 2 · Kinematics

A projectile is projected at

3 0^{\circ}

from horizontal with initial velocity

40 ms^{- 1}

. The velocity of the projectile at

t = 2 s

from the start will be:
(Given

g = 10 m / s^{2}

)

(A) $203 ms^{- 1}$ ✓ Correct
(B) $403 ms^{- 1}$
(C) $20 ms^{- 1}$
(D)Zero

Answer: (A)

Solution

The time to reach maximum height is

t = \frac{u s i n θ}{g} = \frac{40 \times 0.5}{10} = 2

s. At maximum height, the vertical component of velocity is zero, and only the horizontal component remains:

v = u cos θ = 40 cos 3 0^{\circ} = 40 \times \frac{3}{2} = 203

m/s.

JEE Main · 2023 · 11 Apr 2023 · Shift 2 · Single correct option

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