A projectile is projected at 30^ from horizontal with initial vel… — Solution | JEE Physics

MCQEasyJEE 2023Projectile Motion

A projectile is projected at $30^\circ$ from horizontal with initial velocity $40 \, ms^{-1}$ . The velocity of the projectile at $t = 2 \, s$ from the start will be :

(Provided that $g = 10 \, m/s^2$ )

A
Zero
B
$20\sqrt{3} \, ms^{-1}$
C
$40\sqrt{3} \, ms^{-1}$
D
$20 \, ms^{-1}$

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given: Initial speed is $u = 40 \, ms^{-1}$ , angle of projection is $30^\circ$ , time is $t = 2 \, s$ , and $g = 10 \, m/s^2$ .

Find: The velocity of the projectile after $2 \, s$ .

Resolve the initial velocity into components:

U_x = u \cos 30^\circ = 40 \cdot \cos 30^\circ = 40 \cdot \frac{\sqrt{3}}{2} = 20\sqrt{3} \, ms^{-1}

U_y = u \sin 30^\circ = 40 \cdot \sin 30^\circ = 40 \cdot \frac{1}{2} = 20 \, ms^{-1}

The horizontal component remains constant, so

V_x = U_x = 20\sqrt{3} \, ms^{-1}

The vertical component after $t = 2 \, s$ is

V_y = U_y - gt = 20 - 10 \cdot 2 = 0 \, ms^{-1}

Now the resultant velocity is

V = \sqrt{V_x^2 + V_y^2}

Substituting the values:

V = \sqrt{(20\sqrt{3})^2 + 0^2} = \sqrt{1200} = 20\sqrt{3} \, ms^{-1}

Therefore, the velocity of the projectile after $2 \, s$ is $20\sqrt{3} \, ms^{-1}$ . The correct option is B.

Common mistakes

Using the full initial speed $40 \, ms^{-1}$ as the horizontal component is incorrect because projectile motion must first be resolved into components. Use $u\cos 30^\circ$ and $u\sin 30^\circ$ before applying kinematics.
Assuming the horizontal velocity changes due to gravity is wrong because gravity acts only vertically. Keep $V_x$ constant and change only the vertical component with $V_y = U_y - gt$ .
Forgetting that at $t = 2 \, s$ the vertical velocity becomes zero leads to an incorrect resultant speed. First compute $V_y$ , then combine $V_x$ and $V_y$ using the vector magnitude formula.

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Answer

Step-by-step solution

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