Consider a series of steps as shown. A ball is thrown from O. Find the minimum speed of directly jump to 5th step.

Option 1 -

$5 (\sqrt[]{2} + 1) m / s$

Option 2 -

$5 \sqrt[]{2} m / s$

Option 3 -

$5 \sqrt[]{\sqrt[]{2} + 1} m / s$

Option 4 -

$6 \sqrt[]{\sqrt[]{3} + 1} m / s$

2 Views | Posted 2 days ago

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1 Answer

Answered by

alok kumar singh | Contributor-Level 10

2 days ago

Correct Option - 3

Detailed Solution:

$y = x t a n θ - \frac{g x^{2}}{2 v^{2} c o s^{2} θ}$

(2.5, 2.5) must lie on this

-> $1 = t a n θ - \frac{g \times 2.5}{2 v^{2} c o s^{2} θ}$

-> $\frac{25}{2 v^{2} c o s^{2} θ} = t a n θ - 1$

-> $v^{2} = \frac{25}{2} {\frac{1 + t a n^{2} θ}{t a n θ - 1}}$

-> $v_{m i n} = 5 \sqrt[]{\sqrt[]{2} + 1}$

[Happens when tan θ = $\sqrt[]{2}$ + 1]

<math> <mrow> <mi>y</mi> <mo>=</mo> <mi>x</mi> <mi>t</mi> <mi>a</mi> <mi>n</mi> <mi>θ</mi> <mo>−</mo> <mfrac> <mrow> <mi>g</mi> <msup> <mrow> <mi>x</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msup> </mrow> <mrow> <mn>2</mn> <msup> <mrow> <mi>v</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msup> <mi>c</mi> <mi>o</mi> <msup> <mrow> <mi>s</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msup> <mi>θ</mi> </mrow> </mfrac> </mrow> </math> (2.5, 2.5) must lie on this-> <math> <mrow> <mn>1</mn> <mo>=</mo> <mi>t</mi> <mi>a</mi> <mi>n</mi> <mi>θ</mi> <mo>−</mo> <mfrac> <mrow> <mi>g</mi> <mo>×</mo> <mn>2</mn> <mo>.</mo> <mn>5</mn> </mrow> <mrow> <mn>2</mn> <msup> <mrow> <mi>v</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msup> <mi>c</mi> <mi>o</mi> <msup> <mrow> <mi>s</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msup> <mi>θ</mi> </mrow> </mfrac> </mrow> </math> -> <math> <mrow> <mfrac> <mrow> <mn>2</mn> <mn>5</mn> </mrow> <mrow> <mn>2</mn> <msup> <mrow> <mi>v</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msup> <mi>c</mi> <mi>o</mi> <msup> <mrow> <mi>s</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msup> <mi>θ</mi> </mrow> </mfrac> <mo>=</mo> <mi>t</mi> <mi>a</mi> <mi>n</mi> <mi>θ</mi> <mo>−</mo> <mn>1</mn> </mrow> </math> -> <math> <mrow> <msup> <mrow> <mi>v</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msup> <mo>=</mo> <mfrac> <mrow> <mn>2</mn> <mn>5</mn> </mrow> <mrow> <mn>2</mn> </mrow> </mfrac> <mrow> <mo>{</mo> <mrow> <mfrac> <mrow> <mn>1</mn> <mo>+</mo> <mi>t</mi> <mi>a</mi> <msup> <mrow> <mi>n</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msup> <mi>θ</mi> </mrow> <mrow> <mi>t</mi> <mi>a</mi> <mi>n</mi> <mi>θ</mi> <mo>−</mo> <mn>1</mn> </mrow> </mfrac> </mrow> <mo>}</mo> </mrow> </mrow> </math>   -> <math> <mrow> <msub> <mrow> <mi>v</mi> </mrow> <mrow> <mi>m</mi> <mi>i</mi> <mi>n</mi> </mrow> </msub> <mo>=</mo> <mn>5</mn> <mroot> <mrow> <mroot> <mrow> <mn>2</mn> </mrow> <mrow></mrow> </mroot> <mo>+</mo> <mn>1</mn> </mrow> <mrow></mrow> </mroot> </mrow> </math>   [Happens when tan θ = <math> <mrow> <mroot> <mrow> <mn>2</mn> </mrow> <mrow></mrow> </mroot> </mrow> </math>  + 1]

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