Question
Download Solution PDFFor a maximum horizontal range of a projectile, the angle of projection should be
This question was previously asked in
WBMSC Sub-Assistant Engineer (Civil) 05 Jun 2022 Official Paper
Answer (Detailed Solution Below)
Option 2 : 45°
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WBMSC SAE Full Test 1
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Detailed Solution
Download Solution PDFCONCEPT:
- Projectile motion: The horizontal component of velocity (u cosθ ), acceleration (g), and mechanical energy remain a constant while, speed, velocity, the vertical component of velocity (u sinθ), momentum, kinetic energy, and potential energy all changes.
- Velocity and KE are maximum at the point of projection while minimum (but not zero) at the highest point.
- Horizontal range (R): It is the horizontal distance travelled by a body during the time of flight.
EXPLANATION:
- The mathematical expression of the horizontal range is given by:
Horizontal range, \(R = \frac{{u^2\sin 2θ }}{g}\)
For maximum horizontal range:
⇒ Sin 2θ = 1
⇒ 2θ = sin-1(1)
⇒ 2θ = 90°
⇒ θ = 45°Last updated on Jun 14, 2025
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