Explanation

In projectile motion, horizontal component of velocity remains constant throughout the motion.

Let the initial speed of the projectile be u.

Initial horizontal component of velocity:

$$ u_x = u \cos 60^\circ = \frac{u}{2} $$

At the instant when the direction of motion makes 45° with the horizontal, the speed of the projectile is 20 m/s.

At this instant, horizontal and vertical components of velocity are equal in magnitude because:

$$ \tan 45^\circ = \frac{v_y}{v_x} = 1 \Rightarrow v_y = v_x $$

Hence, the speed at that instant is:

$$ v = \sqrt{v_x^2 + v_y^2} = \sqrt{2v_x^2} = v_x\sqrt{2} $$

Given that v = 20 m/s:

$$ 20 = v_x\sqrt{2} $$

$$ v_x = \frac{20}{\sqrt{2}} = 10\sqrt{2} $$

Since horizontal component remains constant:

$$ \frac{u}{2} = 10\sqrt{2} $$

$$ u = 20\sqrt{2}\ \text{m/s} $$

Therefore, the initial speed of the projectile is $20\sqrt{2}$ m/s.

Related JEE Main Questions