Question

an aeroplane takes off at an angle of 60 degree to the horizontal. if the velocity of the plan 200 km/hr calculate it's horizontal and vertical component of it's ​

Answer 1

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$\sf{Here \: \: ν=200 {kmh}^{ - 1}} </p><p>$

$\sf{θ=60{ \degree} } </p><p>$

$\sf{∴ Horizontal \: \: component \: \: </p><p>ν _x</p><p>}$

$\sf{=νcosθ=200cos60{ \degree} </p><p>}$

$\sf{=200× \frac{1}{2} </p><p> }$

$\sf{=100kmh </p><p>^−1</p><p> </p><p>}$

$\sf{Vertical \: \: component</p><p> \: \: ν _y</p><p> }$

$\sf{sinθ=200sin60 </p><p> }$

$\sf{=200× \frac{3}{2} </p><p> </p><p> }$

$\sf{= \frac{100}{3} \: </p><p> kmh </p><p>^−1}$

$\large{\underline{\underline{\red{\bold{hope\: íts \:help\: u\: }}}}}$

Answer 2

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Answer :-

\sf{Here \: \: ν=200 {kmh}^{ - 1}} < /p > < p >Hereν=200kmh

−1

</p><p>

\sf{θ=60{ \degree} } < /p > < p >θ=60°</p><p>

\sf{∴ Horizontal \: \: component \: \: < /p > < p > ν _x < /p > < p > }∴Horizontalcomponent</p><p>ν

x

</p><p>

\sf{=νcosθ=200cos60{ \degree} < /p > < p > }=νcosθ=200cos60°</p><p>

\sf{=200× \frac{1}{2} < /p > < p > }=200×

2

1

</p><p>

\sf{=100kmh < /p > < p > ^−1 < /p > < p > < /p > < p > }=100kmh</p><p>

−

1</p><p></p><p>

\sf{Vertical \: \: component < /p > < p > \: \: ν _y < /p > < p > }Verticalcomponent</p><p>ν

y

</p><p>

\sf{sinθ=200sin60 < /p > < p > }sinθ=200sin60</p><p>

\sf{=200× \frac{3}{2} < /p > < p > < /p > < p > }=200×

2

3

</p><p></p><p>

\sf{= \frac{100}{3} \: < /p > < p > kmh < /p > < p > ^−1}=

3

100

</p><p>kmh</p><p>

−

1

\large{\underline{\underline{\red{\bold{hope\: íts \:help\: u\: }}}}}

hope

ı

ˊ

tshelpu