Class 11th Physics Physics Motion in Plane Projectile Motion

A particle of mass m is projected with speed v at an angle of 30° with the horizontal, find its angular momentum about point of projection when it reaches its maximum height.

Option 1 - <p><span class="mathml" contenteditable="false"> <math> <mrow> <mfrac> <mrow> <mi>m</mi> <msup> <mrow> <mi>v</mi> </mrow> <mrow> <mn>3</mn> </mrow> </msup> </mrow> <mrow> <mn>1</mn> <mn>6</mn> <mi>g</mi> </mrow> </mfrac> </mrow> </math> </span></p>

Option 2 - <p><span class="mathml" contenteditable="false"> <math> <mrow> <mroot> <mrow> <mn>3</mn> </mrow> <mrow></mrow> </mroot> <mfrac> <mrow> <mi>m</mi> <msup> <mrow> <mi>v</mi> </mrow> <mrow> <mn>3</mn> </mrow> </msup> </mrow> <mrow> <mn>1</mn> <mn>6</mn> <mi>g</mi> </mrow> </mfrac> </mrow> </math> </span></p>

Option 3 - <p><span class="mathml" contenteditable="false"> <math> <mrow> <mfrac> <mrow> <mi>m</mi> <msup> <mrow> <mi>v</mi> </mrow> <mrow> <mn>3</mn> </mrow> </msup> </mrow> <mrow> <mn>3</mn> <mi>g</mi> </mrow> </mfrac> </mrow> </math> </span></p>

Option 4 - <p><span class="mathml" contenteditable="false"> <math> <mrow> <mroot> <mrow> <mn>3</mn> </mrow> <mrow></mrow> </mroot> <mfrac> <mrow> <mi>m</mi> <msup> <mrow> <mi>v</mi> </mrow> <mrow> <mn>3</mn> </mrow> </msup> </mrow> <mrow> <mn>8</mn> <mi>g</mi> </mrow> </mfrac> </mrow> </math> </span></p>

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Answered by

Alok Kumar Singh | Contributor-Level 10

4 months ago

Correct Option - 2

Detailed Solution:

Velocity at maximum height = vcoss30°

L = m (vcos30) H

$= m v (\frac{\sqrt[]{3}}{2}) * \frac{v^{2} s i n^{2} 30}{2 g}$

$= \sqrt[]{3} \frac{m v^{3}}{16 g}$

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A ball is projected from ground at angle q with the horizontal. After t = 1 s, it is moving at 45° with the horizontal and after t = 2 second, it is moving horizontally. What is speed of projection of ball? [g = 10 m s^–2]

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The trajectory of a projectile in an vertical plane is y = ax - bx², where a and b are constants and x & y are respectively the horizontal and vertical distances of the projectile from the point of projection. The angle of projection q and the maximum height attained H are respectively given by:

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$y = α x - β x^{2}$

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$y_{m a x} = α \times \frac{α}{2 β} - β \times {(\frac{α}{2 β})}^{2} = \frac{α^{2}}{4 β}$

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A person can throw a ball upto a maximum range of 100m. How high above the ground he can the same ball?

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According to question, we can write

$R = \frac{u^{2} s i n 2 θ}{g} \Rightarrow u^{2} = g R_{m a x}, w h e r e θ = 45 °$

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