Question:
An object is projected with an angle of \( 60^{\circ} \) with horizontal with a velocity V. During the path, when it makes \( 30^{\circ} \) with horizontal, its velocity becomes 10 \( ms^{-1} \), then V is:
An object is projected with an angle of \( 60^{\circ} \) with horizontal with a velocity V. During the path, when it makes \( 30^{\circ} \) with horizontal, its velocity becomes 10 \( ms^{-1} \), then V is:
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The key to projectile motion problems is treating the horizontal component as uniform motion. Since there is no acceleration along the x-axis, \( v_x = \text{constant} \) will quickly unlock the relation between speeds at different angles.
Updated On: Jun 8, 2026
- \( 10\sqrt{3} \, ms^{-1} \)
- \( \sqrt{30} \, ms^{-1} \)
- \( 3\sqrt{10} \, ms^{-1} \)
- \( \sqrt{3} \, ms^{-1} \)
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The Correct Option is A
Solution and Explanation
Concept:
In standard two-dimensional projectile kinematics neglecting atmospheric friction resistance forces, gravity acts purely along the vertical field line. Consequently, the horizontal velocity component remains completely uniform throughout the trajectory:
\[
u_x = v_x \implies V\cos\theta_1 = v\cos\theta_2
\]
Step 1: Equating independent horizontal projection vectors.
Given parameters:
• Initial angle \( \theta_1 = 60^{\circ} \)
• Mid-flight angular tilt \( \theta_2 = 30^{\circ} \)
• Mid-flight speed magnitude \( v = 10 \, ms^{-1} \)
Applying component conservation: \[ V \cos(60^{\circ}) = 10 \cos(30^{\circ}) \]
Step 2: Evaluating trigonometric values.
\[ V \left(\frac{1}{2}\right) = 10 \left(\frac{\sqrt{3}}{2}\right) \] Cancelling out the common factor of 2 from both sides: \[ V = 10\sqrt{3} \, ms^{-1} \]
Step 1: Equating independent horizontal projection vectors.
Given parameters:
• Initial angle \( \theta_1 = 60^{\circ} \)
• Mid-flight angular tilt \( \theta_2 = 30^{\circ} \)
• Mid-flight speed magnitude \( v = 10 \, ms^{-1} \)
Applying component conservation: \[ V \cos(60^{\circ}) = 10 \cos(30^{\circ}) \]
Step 2: Evaluating trigonometric values.
\[ V \left(\frac{1}{2}\right) = 10 \left(\frac{\sqrt{3}}{2}\right) \] Cancelling out the common factor of 2 from both sides: \[ V = 10\sqrt{3} \, ms^{-1} \]
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