Question

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

Answer 1

Answer:

x=40cosθ×2=80cosθ

Applying equation of trajectory ,

20=80cosθ×tanθ−

2×cos

2

θ×40

2

g(80cosθ)

2

20=80sinθ−

2

40

sinθ=

2

1

,θ=30

∘

Range =

g

u

2

sin2θ

=

20

1600×

3

=80

3

=138.5m

Answer 2

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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