Question

A body is projected from ground at an angle of 60 degree from horizontal with speed 60m/s. Find out after how much time it's angle becomes 30 degree from horizontal.​

Answer 1

Answer:

$\frac{6}{ \sqrt{3} } sec$

Explanation:

Given

Angle of projection = 30 °

Speed from horizontal side = 60 m/s

To Find

Time taken by it for reaching 30° from horizontal.

We Know the formula that is

$\tan \theta = \frac{u \sin \theta - gt }{ucos \theta}$

Now On applying Formula

$= > \tan30 \degree = \frac{60 \sin60 \degree - 10t }{60 \cos60 \degree }$

On Solving we get

$= > \frac{1}{ \sqrt{3} } = \frac{60 \times \frac{ \sqrt{3} }{2 } - 10t }{ \cancel60 \times \frac{1}{ \cancel2} } \\ = > \frac{1}{ \sqrt{3} } \times \cancel60 \times \frac{1}{ \cancel2} = \cancel60 \times \frac{ \sqrt{3} }{ \cancel2} - 10t \\ = > 10t = 30 \sqrt{3} - \frac{30}{ \sqrt{3} } \\ = > 10t = \frac{90 - 30}{ \sqrt{3} } \\ = > \cancel10t = \frac{ \cancel60}{ \sqrt{3} } \\ = > t = {\frac{6}{ \sqrt{3} } sec}$

Hence

$After \: \frac{6}{ \sqrt{3} } \: its \: angle \: will \\ become \: 30\degree \: from \: horizontal.$

Answer 2

$\bold {\huge {Your ~answer :-}}$

$\bf\huge\frac{6}{ \sqrt{3} } sec$

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As we Know That

$\tan \theta = \frac{u \sin \theta - gt }{ucos \theta}$

Now

$= > \tan30 \degree = \frac{60 \sin60 \degree - 10t }{60 \cos60 \degree }$

By solving

Above

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