Question

The velocity of a projectile when it is half of the maximum height is√5/2 times of velocity at its highest point. The angle of projection is​

Answer 1

Answer:

Hey mate here your answer

Explanation:

Let us consider that the maximum height covered by the projectile is H and the angle of projection is θ.

It is given that,

v_{H}=\sqrt{\frac{2}{5}} v_{H / 2}v

H

=

5

2

v

H/2

So the maximum height of a projectile motion can be found as

H=\frac{u^{2} \sin ^{2} \theta}{2 g} \quad \rightarrow(1)H=

2g

u

2

sin

2

θ

→(1)

In the above equation, u is the initial velocity and g is the acceleration due to gravity.

We know that velocity at maximum height is v_{H}=u \cos \thetav

H

=ucosθ

Squaring on both sides, we get v_{H}^{2}=u^{2} \cos ^{2} \theta \quad \rightarrow(2)v

H

2

=u

2

cos

2

θ→(2)

From Newton’s second law of motion:

v^{2}=u^{2}-2 a s \quad \rightarrow(3)v

2

=u

2

−2as→(3)

We can substitute v=v_{H / 2}v=v

H/2

, as the velocity at half the maximum height and displacement (s) can be replaced by half the maximum height, i.e., \mathrm{s}=\mathrm{H} / 2\ in\ (3)s=H/2 in (3)

v_{H / 2}^{2}=u^{2}-2 g\left(\frac{H}{2}\right) \quad \rightarrow(4)v

H/2

2

=u

2

−2g(

2

H

)→(4)

The H in eqn (4) can be replaced with the value of H in eqn (1)

v_{H / 2}^{2}=u^{2}-2 g\left(\frac{\frac{u^{2} \sin ^{2} \theta}{2 g}}{2}\right) \quad \rightarrow(5)v

H/2

2

=u

2

−2g(

2

2g

u

2

sin

2

θ

)→(5)

v_{H / 2}^{2}=u^{2}-2 g\left(\frac{u^{2} \sin ^{2} \theta}{4 g}\right) \quad \rightarrow(6)v

H/2

2

=u

2

−2g(

4g

u

2

sin

2

θ

)→(6)

It is given that v_{H}=\sqrt{\frac{2}{5}} v_{H / 2}v

H

=

5

2

v

H/2

So squaring on both sides, we get

v_{H}^{2}=\frac{2}{5} v_{H / 2}^{2} \rightarrow(7)v

H

2

=

5

2

v

H/2

2

→(7)

So, substitute eqn (2) and eqn (6) in eqn (7)

u^{2} \cos ^{2} \theta=\frac{2}{5}\left[u^{2}-2 g\left(\frac{u^{2} \sin ^{2} \theta}{4 g}\right)\right]u

2

cos

2

θ=

5

2

[u

2

−2g(

4g

u

2

sin

2

θ

)]

(We know that, \cos ^{2} \theta+\sin ^{2} \theta=1cos

2

θ+sin

2

θ=1 )

\therefore u^{2}\left(1-\sin ^{2} \theta\right)=\frac{2 u^{2}}{5}\left[1-2\left(\frac{\sin ^{2} \theta}{4}\right)\right]∴u

2

(1−sin

2

θ)=

5

2u

2

[1−2(

4

sin

2

θ

)]

\left(1-\sin ^{2} \theta\right)=\frac{2}{5}\left[1-\left(\frac{\sin ^{2} \theta}{2}\right)\right](1−sin

2

θ)=

5

2

[1−(

2

sin

2

θ

)]

1-\sin ^{2} \theta=\frac{2}{5}-\frac{1}{5} \sin ^{2} \theta1−sin

2

θ=

5

2

−

5

1

sin

2

θ

\frac{1}{5} \sin ^{2} \theta-\sin ^{2} \theta=\frac{2}{5}-1

5

1

sin

2

θ−sin

2

θ=

5

2

−1

-\frac{4}{5} \sin ^{2} \theta=-\frac{3}{5}−

5

4

sin

2

θ=−

5

3

\sin ^{2} \theta=\frac{3}{5} \times \frac{5}{4}sin

2

θ=

5

3

×

4

5

\sin ^{2} \theta=\frac{3}{4}sin

2

θ=

4

3

\therefore \sin \theta=\frac{\sqrt{3}}{2}∴sinθ=

2

3

\bold{\therefore \theta=\sin ^{-1}\left(\frac{\sqrt{3}}{2}\right)=60^{\circ}}∴θ=sin

−1

(

2

3

)=60

∘

Thus the angle of projection is 60°.

Answer 2

The angle of projection is :

θ = $sin^{-1} \sqrt{2-\sqrt{5} }$

Given:

The velocity of a projectile when it is half of the maximum height is√5/2 times of velocity at its highest point.

To Find:

The angle of projection.

Solution:

Time taken to reach max height = usinθ/g

Max height = $u^{2} sin^{2}$θ/2g

Half the max height  = $u^{2} sin^{2}$θ/4g  

Vertical velocity at half max height = usinθ/$\sqrt{2}$

Velocity at half the max-height

= $\sqrt{u^{2} cos^{2} + u^{2} sin^{2}/2 }$

=> ucosθ = $\sqrt{5}$/2 $\sqrt{u^{2} cos^{2} + u^{2} sin^{2}/2 }$

=> $u^{2} cos^{2}$θ  =  $\sqrt{5}$/2 ($u^{2}$)(1 - $sin^{2}$θ  + $sin^{2}/2$)

=> 2( 1 - $sin^{2}$θ ) =  $\sqrt{5}$ - $sin^{2}$θ

=> sinθ = $\sqrt{2-\sqrt{5} }$

Hence , the angle of projection is :

θ = $sin^{-1} \sqrt{2-\sqrt{5} }$

#SPJ3

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