NEET 2022 Physics Projectile Motion MCQ Question

Chapter: Motion in a Plane

Class: 11 Physics Topic: Projectile Motion Difficulty: 🟢 Easy

✅ Ans: D — $5\sqrt3\,m/s$

Given:

$$u=10\,m/s$$

Angle with vertical:

$$60^\circ$$

Therefore, angle with horizontal:

$$\theta=90^\circ-60^\circ=30^\circ$$

At the highest point:

$$v_y=0$$

but horizontal velocity remains constant:

$$v_x=u\cos30^\circ$$ $$v_x=10\times\frac{\sqrt3}{2}$$ $$\boxed{v_x=5\sqrt3\,m/s}$$

Hence the speed at the highest point is:

$$\boxed{5\sqrt3\,m/s}$$

❌ Why other options are wrong?

• A: $5\,m/s$ ❌ This is the vertical component initially:

  $$u_y=10\sin30^\circ=5\,m/s$$

• B: $10\,m/s$ ❌ Initial speed, not the speed at the highest point.

• C: Zero ❌ Only the vertical component becomes zero; horizontal velocity remains.

• D: $5\sqrt3\,m/s$ ✅ Correct.

📌 NCERT Concept

At the highest point of projectile motion:

$$\boxed{v_y=0,\qquad v_x=u\cos\theta}$$

Horizontal velocity remains unchanged because horizontal acceleration is zero.

🧠 NEET Trick

If angle is given with vertical:

$$\boxed{v_{\text{highest}}=u\sin60^\circ}$$ $$=10\frac{\sqrt3}{2} =\boxed{5\sqrt3\,m/s}$$

Final Answer: $\boxed{D}$.