A gun can fire shells with maximum speed vo and the maximum horizontal range that can be achieved is R= <math> <mfrac> <mrow> <mrow> <msubsup> <mrow> <mrow> <mi mathvariant="bold-italic">v</mi> </mrow> </mrow> <mrow> <mrow> <mi mathvariant="bold-italic">o</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msubsup> </mrow> </mrow> <mrow> <mrow> <mi mathvariant="bold-italic">g</mi> </mrow> </mrow> </mfrac> </math> .If a target farther away by distance ∆x (beyond R) has to be hit with the same gun (Fig), show that it could be achieved by raising the gun to a height at least h = <math> <mo>∆</mo> <mi mathvariant="bold-italic">x</mi> <mfenced open="[" close="]" separators="|"> <mrow> <mrow> <mn>1</mn> <mo>+</mo> <mfrac> <mrow> <mrow> <mo>∆</mo> <mi mathvariant="bold-italic">x</mi> </mrow> </mrow> <mrow> <mrow> <mi mathvariant="bold-italic">R</mi> </mrow> </mrow> </mfrac> </mrow> </mrow> </mfenced> <mo>.</mo> </math>

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Moving Charges and Magnetism Physics Ncert Solutions Class 12th Physics Class 12th Physics NCERT Exemplar Solutions Class 12th Chapter Four

A gun can fire shells with maximum speed vo and the maximum horizontal range that can be achieved is R= $\frac{v_{o}^{2}}{g}$ .If a target farther away by distance ∆x (beyond R) has to be hit with the same gun (Fig), show that it could be achieved by raising the gun to a height at least h = $∆ x [1 + \frac{∆ x}{R}] .$

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alok kumar singh | Contributor-Level 10

4 months ago

This is a Long Answer Type Questions as classified in NCERT Exemplar

Explanation – target T is at horizontal distance x= R+ $? x$ and between point of projection y= -h

Maximum horizontal range R= $\frac{v_{o}^{2}}{g} θ = 45$ …………1

Horizontal component of initial velocity = V_ocos $θ$

Vertical component of initial velocity = -V_osin $θ$

So h = (-V_osin $θ$ )t + $\frac{1}{2} g t$ ²………….2

R+ $? x$ = V_ocos $θ \times t$

So t= $\frac{R + ? x}{v_{o} c o s θ}$

Substituting value of t in 2 we get

So h = (-V₀sin $θ$ ) $(\frac{R + ? x}{v_{o} c o s θ}) + \frac{1}{2} g ({\frac{R + ? x}{v_{o} c o s θ})}^{2}$

H = -(R+ $? x$ )tan $θ$ + $\frac{1}{2} g (\frac{({R + ? x)}^{2}}{v_{o}^{2} {c o s}^{2} θ})$

$θ = 45$ , h = -(R+ $? x$ )tan $45$ + $\frac{1}{2} g (\frac{({R + ? x)}^{2}}{v_{o}^{2} {c o s}^{2} 45})$

So h = -(R+ $? x$ ) $1$ + $\frac{1}{2} g (\frac{({R + ? x)}^{2}}{v_{o}^{2} (\frac{1}{2})})$

So h = -(R+ $? x$ )+ $\frac{{(R + ? x)}^{2}}{R}$

So h = -R- $? x$ +(R+ $\frac{{? x}^{2}}{R} + 2 ? x$ )

h= $? x + \frac{{? x}^{2}}{R}$

This is a Long Answer Type Questions as classified in NCERT ExemplarExplanation – target T is at horizontal distance x= R+ <math> <mo>?</mo> <mi>x</mi> </math> and between point of projection y= -hMaximum horizontal range R= <math> <mfrac> <mrow> <mrow> <msubsup> <mrow> <mrow> <mi>v</mi> </mrow> </mrow> <mrow> <mrow> <mi>o</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msubsup> </mrow> </mrow> <mrow> <mrow> <mi>g</mi> </mrow> </mrow> </mfrac> <mi> </mi> <mi> </mi> <mi>θ</mi> <mo>=</mo> <mn>45</mn> </math> …………1Horizontal component of initial velocity = Vocos <math> <mi>θ</mi> </math> Vertical component of initial velocity = -Vosin <math> <mi>θ</mi> </math> So h = (-Vosin <math> <mi>θ</mi> </math> )t + <math> <mfrac> <mrow> <mrow> <mn>1</mn> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </mfrac> <mi>g</mi> <mi>t</mi> </math> 2………….2R+ <math> <mo>?</mo> <mi>x</mi> </math>  = Vocos <math> <mi>θ</mi> <mo>×</mo> <mi>t</mi> </math> So t= <math> <mfrac> <mrow> <mrow> <mi>R</mi> <mo>+</mo> <mo>?</mo> <mi>x</mi> </mrow> </mrow> <mrow> <mrow> <msub> <mrow> <mrow> <mi>v</mi> </mrow> </mrow> <mrow> <mrow> <mi>o</mi> </mrow> </mrow> </msub> <mi>c</mi> <mi>o</mi> <mi>s</mi> <mi>θ</mi> </mrow> </mrow> </mfrac> </math> Substituting value of t in 2 we getSo h = (-V0sin <math> <mi>θ</mi> </math> ) <math> <mfenced separators="|"> <mrow> <mrow> <mi mathvariant="normal"> </mi> <mfrac> <mrow> <mrow> <mi>R</mi> <mo>+</mo> <mo>?</mo> <mi>x</mi> </mrow> </mrow> <mrow> <mrow> <msub> <mrow> <mrow> <mi>v</mi> </mrow> </mrow> <mrow> <mrow> <mi>o</mi> </mrow> </mrow> </msub> <mi>c</mi> <mi>o</mi> <mi>s</mi> <mi>θ</mi> </mrow> </mrow> </mfrac> </mrow> </mrow> </mfenced> <mo>+</mo> <mfrac> <mrow> <mrow> <mn>1</mn> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </mfrac> <mi>g</mi> <mo>(</mo> <msup> <mrow> <mrow> <mi mathvariant="normal"> </mi> <mfrac> <mrow> <mrow> <mi>R</mi> <mo>+</mo> <mo>?</mo> <mi>x</mi> </mrow> </mrow> <mrow> <mrow> <msub> <mrow> <mrow> <mi>v</mi> </mrow> </mrow> <mrow> <mrow> <mi>o</mi> </mrow> </mrow> </msub> <mi>c</mi> <mi>o</mi> <mi>s</mi> <mi>θ</mi> </mrow> </mrow> </mfrac> <mo>)</mo> <mi> </mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> </math> H = -(R+ <math> <mo>?</mo> <mi>x</mi> </math> )tan <math> <mi>θ</mi> </math> + <math> <mfrac> <mrow> <mrow> <mn>1</mn> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </mfrac> <mi>g</mi> <mfenced separators="|"> <mrow> <mrow> <mfrac> <mrow> <mrow> <mo>(</mo> <msup> <mrow> <mrow> <mi>R</mi> <mo>+</mo> <mo>?</mo> <mi>x</mi> <mo>)</mo> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> </mrow> </mrow> <mrow> <mrow> <msubsup> <mrow> <mrow> <mi>v</mi> </mrow> </mrow> <mrow> <mrow> <mi>o</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msubsup> <msup> <mrow> <mrow> <mi>c</mi> <mi>o</mi> <mi>s</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> <mi>θ</mi> </mrow> </mrow> </mfrac> </mrow> </mrow> </mfenced> </math> <math> <mi>θ</mi> <mo>=</mo> <mn>45</mn> </math> ,  h = -(R+ <math> <mo>?</mo> <mi>x</mi> </math> )tan <math> <mn>45</mn> </math> + <math> <mfrac> <mrow> <mrow> <mn>1</mn> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </mfrac> <mi>g</mi> <mfenced separators="|"> <mrow> <mrow> <mfrac> <mrow> <mrow> <mo>(</mo> <msup> <mrow> <mrow> <mi>R</mi> <mo>+</mo> <mo>?</mo> <mi>x</mi> <mo>)</mo> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> </mrow> </mrow> <mrow> <mrow> <msubsup> <mrow> <mrow> <mi>v</mi> </mrow> </mrow> <mrow> <mrow> <mi>o</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msubsup> <msup> <mrow> <mrow> <mi>c</mi> <mi>o</mi> <mi>s</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> <mn>45</mn> </mrow> </mrow> </mfrac> </mrow> </mrow> </mfenced> </math> So h = -(R+ <math> <mo>?</mo> <mi>x</mi> </math> ) <math> <mn>1</mn> </math> + <math> <mfrac> <mrow> <mrow> <mn>1</mn> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </mfrac> <mi>g</mi> <mfenced separators="|"> <mrow> <mrow> <mfrac> <mrow> <mrow> <mo>(</mo> <msup> <mrow> <mrow> <mi>R</mi> <mo>+</mo> <mo>?</mo> <mi>x</mi> <mo>)</mo> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> </mrow> </mrow> <mrow> <mrow> <msubsup> <mrow> <mrow> <mi>v</mi> </mrow> </mrow> <mrow> <mrow> <mi>o</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msubsup> <mfenced separators="|"> <mrow> <mrow> <mfrac> <mrow> <mrow> <mn>1</mn> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </mfrac> </mrow> </mrow> </mfenced> </mrow> </mrow> </mfrac> </mrow> </mrow> </mfenced> </math> So h = -(R+ <math> <mo>?</mo> <mi>x</mi> </math> )+ <math> <mfrac> <mrow> <mrow> <msup> <mrow> <mrow> <mo>(</mo> <mi>R</mi> <mo>+</mo> <mo>?</mo> <mi>x</mi> <mo>)</mo> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> </mrow> </mrow> <mrow> <mrow> <mi>R</mi> </mrow> </mrow> </mfrac> </math> So h = -R- <math> <mo>?</mo> <mi>x</mi> </math> +(R+ <math> <mfrac> <mrow> <mrow> <msup> <mrow> <mrow> <mo>?</mo> <mi>x</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> </mrow> </mrow> <mrow> <mrow> <mi>R</mi> </mrow> </mrow> </mfrac> <mo>+</mo> <mn>2</mn> <mo>?</mo> <mi>x</mi> </math> ) h= <math> <mo>?</mo> <mi>x</mi> <mo>+</mo> <mfrac> <mrow> <mrow> <msup> <mrow> <mrow> <mo>?</mo> <mi>x</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> </mrow> </mrow> <mrow> <mrow> <mi>R</mi> </mrow> </mrow> </mfrac> </math>

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A gun can fire shells with maximum speed vo and the maximum horizontal range that can be achieved is R= v o 2 g .If a target farther away by distance ∆x (beyond R) has to be hit with the same gun (Fig), show that it could be achieved by raising the gun to a height at least h = ∆ x 1 + ∆ x R .

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