Question

Solve the given question....​

Answer 1

↑↑↑↑↑↑↑DIAGRAM↑↑↑↑↑↑↑

STEP BY STEP EXPLANATION:

→FOR THE FOOTBALL:

• Initial speed,u=20 m/s

• Angle,∅=45°(Since, range is maximum)

• Acc due to gravity,g=10 m/s²

Therefore, using,,

$\sf{\boxed{R_{Max}=\frac{u^{2}}{g}}}$

We get,,

$\sf{R_{Max}=\frac{20^{2}}{10}}$

$\sf{\implies{R_{Max}=\frac{400}{10}}}$

$\sf{\implies{\boxed{R_{Max}=40 m}}}$

Now,using,,

$\sf{\boxed{T=\frac{2u\sin\theta}{g}}}$

We get,,

Time taken,,$\sf{T=\frac{2\times{20}\sin{45°}}{10}}$

$\sf{\implies{T=\frac{40}{10\times{\sqrt{2}}}}}$

$\sf{\implies{T=2\sqrt{2} s}}$

→FOR THE SECOND PLAYER:

• Distance that the second player has to travel,,S=40-24=16m

• The time in which he has to cover the distance of 16m,t=T=2√2 s

• To find, speed of the second player,v=??

Using,,

$\sf{v=\frac{S}{t}}$

We get,,

$\sf{v=\frac{16}{2\sqrt{2}}}$

$\sf{\implies {\boxed{v=4\sqrt{2} m/s^{-1}}Ans..}}$

$\rule{200}2$

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Answer 2

Answer:

• Speed (v) of the second player is 4√(2) m/s

Given:

1. Speed of football (u) = 20 m/s.
2. Distance between two players (s) = 24 m.
3. Acceleration due to gravity (g) = 10 m/s²

Explanation:

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Diagram:-

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Let the Player who kicks the football (First player) be at position A and second player be at position B.

Now in this question firstly we need to find that how much distance the football travels and after what time it will land the ground i.e. Range and Time of flight.

Range:

From the Formula we know,

$\large\bigstar\;\underline{\boxed{\sf R=\dfrac{u^{2}\sin 2\theta}{g}}}$

Here,

• R Denotes Range.
• u Denotes initial velocity.
• g Denotes acceleration due to gravity.

Substituting the values,

As it is given that the Range is maximum, therefore θ = 45°,

$\displaystyle\longrightarrow\sf R=\dfrac{\bigg[20\bigg]^{2}\times \sin (2\times 45)^{\circ}}{10}\\\\\\\longrightarrow\sf R=\dfrac{400\times \sin 90^{\circ}}{10}\\\\\\\longrightarrow\sf R=\dfrac{400\times 1}{10} \ \ \ \because\Bigg[\sin 90^{\circ}=1\Bigg]\\\\\\\longrightarrow\sf R=\dfrac{400}{10}\\\\\\\longrightarrow\sf R=40\\\\\\\longrightarrow \underline{\boxed{\sf R=40\;m}}$

∴ Range (R) is 40 meters.

Time of Flight:-

From the formula we know,

$\large\bigstar\;\underline{\boxed{\sf T=\dfrac{2u\sin \theta}{g}}}$

Here,

• T Denotes time of flight.
• u Denotes initial velocity.
• g Denotes acceleration due to gravity.

Substituting the values,

$\displaystyle\longrightarrow\sf T=\dfrac{2\times 20\times \sin 45^{\circ}}{10}\\\\\\\longrightarrow\sf T=\dfrac{40\times 1/\sqrt{2}}{10}\ \ \ \because \Bigg[\sin 45^{\circ}=\dfrac{1}{\sqrt{2}}\Bigg]\\\\\\\longrightarrow\sf T=\dfrac{4}{\sqrt{2}}\\\\\\\longrightarrow\sf T=\dfrac{4\times \sqrt{2}}{\sqrt{2}\times \sqrt{2}}\\\\\\\longrightarrow\sf T=\dfrac{4\sqrt{2}}{2}\\\\\\\longrightarrow\sf T=2\sqrt{2}\\\\\\\longrightarrow\underline{\boxed{\sf T=2\sqrt{2} \; s}}$

∴  Time of flight (T) is 2√(2) seconds.

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As we know the distance (s) between the players is 24 m, so to catch the ball second player has to run in positive x - direction.

Therefor he need to run a distance of,

⇒ d = R - s

Here,

• R Denotes range.
• s Denotes distance between players.
• d Denotes the distance travelled by 2nd player.

⇒ d = 40 - 24

⇒ d = 16

⇒ d = 16 m

∴ Second player travels a Distance of 16 m.

Now, he has to travel this distance in the given time period (or) Time of flight.

From the formula we know,

$\large\bigstar\;\underline{\boxed{\sf V=\dfrac{d}{T}}}$

Here,

• V Denotes speed of 2nd player.
• d Denotes distance travelled by 2nd player.
• T Denotes Time of flight.

Substituting the values,

$\displaystyle\longrightarrow\sf V=\dfrac{16}{2\;\sqrt{2}}\\\\\\\longrightarrow\sf V=\dfrac{8}{\sqrt{2}}\\\\\\\longrightarrow\sf V=\dfrac{8\times \sqrt{2}}{\sqrt{2}\times \sqrt{2}}\\\\\\\longrightarrow\sf V=\dfrac{8\sqrt{2}}{2}\\\\\\\longrightarrow\sf V=4\sqrt{2} \\\\\\\longrightarrow\large{\underline{\boxed{\red{\sf V=4\sqrt{2}\;m/s}}}}$

∴ Speed (v) of the second player is 4√(2) m/s.

$\rule{300}{1.5}$