Motion in a Plane Physics NCERT Exemplar Solutions Class 12th Chapter Four Class 12th Physics

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

Option 1 - <p><span class="mathml" contenteditable="false"> <math> <mrow> <mi>t</mi> <mi>a</mi> <msup> <mrow> <mi>n</mi> </mrow> <mrow> <mo>−</mo> <mn>1</mn> </mrow> </msup> <mi>α</mi> <mo>,</mo> <mfrac> <mrow> <mn>4</mn> <msup> <mrow> <mi>α</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msup> </mrow> <mrow> <mi>β</mi> </mrow> </mfrac> </mrow> </math> </span></p>

Option 2 - <p><span class="mathml" contenteditable="false"> <math> <mrow> <mi>t</mi> <mi>a</mi> <msup> <mrow> <mi>n</mi> </mrow> <mrow> <mo>−</mo> <mn>1</mn> </mrow> </msup> <mi>α</mi> <mo>,</mo> <mfrac> <mrow> <msup> <mrow> <mi>α</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msup> </mrow> <mrow> <mn>4</mn> <mi>β</mi> </mrow> </mfrac> </mrow> </math> </span></p>

Option 3 - <p><span class="mathml" contenteditable="false"> <math> <mrow> <mi>t</mi> <mi>a</mi> <msup> <mrow> <mi>n</mi> </mrow> <mrow> <mo>−</mo> <mn>1</mn> </mrow> </msup> <mi>β</mi> <mo>,</mo> <mfrac> <mrow> <msup> <mrow> <mi>α</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msup> </mrow> <mrow> <mn>2</mn> <mi>β</mi> </mrow> </mfrac> </mrow> </math> </span></p>

Option 4 - <p><span class="mathml" contenteditable="false"> <math> <mrow> <mi>t</mi> <mi>a</mi> <msup> <mrow> <mi>n</mi> </mrow> <mrow> <mo>−</mo> <mn>1</mn> </mrow> </msup> <mrow> <mo>(</mo> <mrow> <mfrac> <mrow> <mi>β</mi> </mrow> <mrow> <mi>α</mi> </mrow> </mfrac> </mrow> <mo>)</mo> </mrow> <mo>,</mo> <mfrac> <mrow> <msup> <mrow> <mi>α</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msup> </mrow> <mrow> <mi>β</mi> </mrow> </mfrac> </mrow> </math> </span></p>

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Answered by

Payal Gupta | Contributor-Level 10

6 months ago

Correct Option - 2

Detailed Solution:

$y = α x - β x^{2}$

$\Rightarrow \frac{d y}{d x} = α - 2 β x = 0$

$\Rightarrow x = \frac{α}{2 β}$

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

Range = $2 x = \frac{α}{β} = \frac{2 u^{2} s i n θ . c o s θ}{g}$

On comparing with

y = x tan θ- $\frac{g x^{2}}{2 u^{2} c o s^{2} θ}$

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Physics NCERT Exemplar Solutions Class 12th Chapter Four 2025

Physics NCERT Exemplar Solutions Class 12th Chapter Four 2025

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