Physics, asked by pateldiya2510, 7 months ago

A body is projected with a velocity of 40m/s at an angle 30° with the horizontal. Find displacement of the body in the first two seconds.



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Answers

Answered by parthchavan40
1

Answer:

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Answered by sonuvuce
6

The displacement in first 2 seconds is 20√13 m

Explanation:

Given:

Initial velocity of projectile v = 40 m/s

Angle of inclination \theta=30^\circ

To find out:

The displacement of the body in first two seconds

Solution:

The horizontal component of the velocity

=v\cos\theta=40\cos 30^\circ=40\times\frac{\sqrt{3}}{2}=20\sqrt{3} m/s

The vertical component of the velocity

=v\sin\theta=40\sin 30^\circ=40\times\frac{1}{2}=20 m/s

After two seconds

The horizontal displacement of the particle

x=20\sqrt{3}\times 2

x=40\sqrt{3} m

Vertical displacement of the particle

y=20\times 2-\frac{1}{2}\times 10\times 2^2

\implies y=40-20

\implies y=20 m

Therefore, the displacement of the projectile

s=\sqrt{x^2+y^2}

\implies s=\sqrt{(40\sqrt{3})^2+20^2}

\implies s=\sqrt{4800+400}

\implies s=\sqrt{5200}

\implies s=\sqrt{13\times 400}

\implies s=20\sqrt{13} m

Hope this answer is helpful.

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