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SS2: PHYSICS - 1ST TERM

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  1. Scalars & Vectors | Week 1
    5 Topics
    |
    1 Quiz
  2. Equations of Motion | Week 2
    3 Topics
    |
    1 Quiz
  3. Projectile | Week 3
    5 Topics
  4. Equilibrium of Forces I | Week 4
    4 Topics
  5. Equilibrium of Forces II | Week 5
    4 Topics
  6. Stability of a Body | Week 6
    4 Topics
    |
    1 Quiz
  7. Simple Harmonic Motion (SHM) | Week 7
    4 Topics
  8. Speed, Velocity & Acceleration & Energy of Simple Harmonic Motion | Week 8
    5 Topics
    |
    1 Quiz
  9. Linear Momentum | Week 9
    6 Topics
    |
    1 Quiz
  10. Mechanical Energy & Machines | Week 10
    2 Topics
    |
    1 Quiz



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Range of a projectile is the horizontal distance covered by a projectile from the point of projection to the point where it strikes the projection plane again.

If the time taken to cover the horizontal distance is T, and the horizontal component of the velocity is U cos θ , then

Range = horizontal velocity x Time

R = U cosθ x T

 R= U cosθ x \( \frac {2U Sinθ}{g} \)

R = \( \frac {2U^2 Sinθcosθ}{g} \)

From trigonometry, 2sinθcosθ = 2θ

Then R = \( \frac {U^2 Sin2θ}{g} \)

When θ = 45°, Range is maximum = 2 x 45° = 90°

Then, R = \( \frac {U^2 Sin90^o}{g} \)

Rmax = \( \frac{U^2}{g}\)

Hence, maximum horizontal range can be achieved when θ is 45°.

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