Question

If a projectile is projected with speed \(v\) at an angle \(60^\circ\) with the horizontal, find the ratio of speed at the highest point (\(v_B\)) to the speed at the final point (\(v_C\)).

• A. \(3:4\)
• B. \(1:3\)
• C. \(1:2\)
• D. \(1:12\)

Hint:

At the highest point of a projectile, the speed is purely horizontal and equals \(v\cos\theta\).

Answer 1

The Correct Option is C

Alternative Method (Energy & Symmetry Approach):

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In projectile motion without air resistance, the total mechanical energy of the projectile remains conserved throughout the motion.

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Step 1: Compare speeds using conservation of energy

Let the projectile be thrown with initial speed v. Since the projectile lands at the same vertical level from which it was projected, the loss and gain of gravitational potential energy cancel out.

Hence, the speed at the final point is equal to the initial speed:

v_C = v

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Step 2: Speed at the highest point

At the highest point of the trajectory, the vertical component of velocity becomes zero. The speed at this point is therefore entirely due to the horizontal component of velocity.

The horizontal component of velocity remains constant throughout the motion and is given by:

v_B = v cos 60° = v / 2

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Step 3: Find the required ratio

v_B : v_C = (v / 2) : v

= 1 : 2

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Final Answer:

v_B : v_C = 1 : 2