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

Concept:

In projectile motion (neglecting air resistance), horizontal velocity remains constant.
At the highest point, vertical component of velocity becomes zero.
If a projectile lands at the same level from which it was projected, the speed at the final point equals the initial speed.
Step 1: Resolve the initial velocity. Initial speed \(= v\) Horizontal component: \[ v_x = v\cos 60^\circ = \frac{v}{2} \] Vertical component: \[ v_y = v\sin 60^\circ \]
Step 2: Speed at the highest point (\(v_B\)). At the highest point: \[ v_y = 0 \] Only horizontal component remains: \[ v_B = v_x = \frac{v}{2} \]
Step 3: Speed at the final point (\(v_C\)). Since the projectile lands at the same horizontal level: \[ v_C = v \]
Step 4: Required ratio. \[ \frac{v_B}{v_C} = \frac{\frac{v}{2}}{v} = \frac{1}{2} \] \[ \boxed{v_B : v_C = 1 : 2} \]