tex] with the horizontal. ...

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Answer

Angle of projection = 60°

Given

A projectile is thrown with some initial velocity at an angle α to the horizontal. Its velocity when it is at the highest point is √2/5 times the velocity when it is at height half of the maximum height

To Find

The angle of Projection α with the horizontal

Solution

In horizontal projection ,

$\rm v_y=0\\\\\rm So, v_x=u_x+a_xt\\\\\to\ \rm v_x=u_x\ [\; \because\ a_x=0\ ]\\\\\to\ \rm v_x=ucos\theta\\\\\rm Now,v_1=\sqrt{v_x^2+v_y^2}\\\\\to\ \rm v_1=ucos\theta...(1)$

A/c , " velocity when it is at the highest point is √2/5 times the velocity "

Apply 3rd equation of motion ,

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Given that ,

$\rm \dfrac{v_1}{v_2}=\sqrt{\dfrac{2}{5}}\\\\\to\ \rm \dfrac{ucos\theta}{\sqrt{u^2cos^2\theta+\dfrac{u^2sin^2\theta}{2}}}=\sqrt{\dfrac{2}{5}}\\\\\to\ \rm \dfrac{u^2cos^2\theta}{u^2cos^2\theta+\dfrac{u^2sin^2\theta}{2}}=\dfrac{2}{5}\\\\\rm \to\ \dfrac{cos^2\theta}{cos^2\theta+\dfrac{sin^2\theta}{2}}=\dfrac{2}{5}\\\\\rm \to\ \dfrac{1}{1+\dfrac{tan^2\theta}{2}}=\dfrac{2}{5}\\\\\rm \to\ \dfrac{2}{2+tan^2\theta}=\dfrac{2}{2+3}\\\\\rm \to\ tan^2\theta=3\\\\\rm\to\ tan\theta=\sqrt{3}\\\\\rm\to\ \theta=60^0$

So , Angle of projection , θ = α = 60°

rocky200216

$\large\mathcal{\underbrace{\red{SOLUTION:-}}}$

GIVEN :-

A projectile is thrown with some initial Velocity at an angle $\rm{\alpha}$ to the horizontal .

It's velocity when it is at the highest point is $\rm{\sqrt{\dfrac{2}{5}}}$ times the velocity when it is at height half of the maximum height .

FORMULA :-

$\checkmark\:\rm{\purple{\boxed{Time\:taken\:to\:reach\:maximum\:height\:=\:\dfrac{u^2\:\sin^2\theta}{2g}\:}}}$

$\checkmark\:\rm{\blue{\boxed{Time\:taken\:to\:reach\:Half\:of\:the\:maximum\:height\:=\:\dfrac{u^2\:\sin^2\theta}{4g}\:}}}$

CALCULATION :-

✍️ Let, initial velocity of the projectile motion is ‘u’ .

Here, $\rm{\underline{\theta\:=\:\alpha}}$ .

✍️ We have know that, at the maximum height of the projectile motion

vertical velocity = 0 m/s

And initial velocity = ucos $\rm{\alpha}$

✍️ Now, vertical velocity at the half of the maximum height is

$\rm{\:=\:\sqrt{u^2\:\sin^2\alpha\:-\:2g\:({\dfrac{u\sin{\alpha}}{2g}})\:}}$

$\rm{\underline{\:=\:\dfrac{u\sin{\alpha}}{\sqrt{2}}\:}}$

✴️ So, net velocity at half of the maximum height is

$\rm{=\:\sqrt{u^2\:\cos^2\alpha\:+\:\dfrac{u^2\:\sin^2\alpha}{2}\:}}$

✍️ According to the question,

$u \cos \alpha = \sqrt{ \frac{2}{5} } \sqrt{ {u}^{2} { \cos^2 \alpha } + \frac{ {u}^{2} { \sin^2 \alpha } }{2} } \\ \\ = > {u}^{2} { \cos^2 \alpha } = \frac{2}{5} {u}^{2} ( { \cos^2 \alpha } + \frac{ { \sin^2 \alpha } }{2} ) \\ \\ = > {u}^{2} { \cos^2 \alpha } = \frac{2}{5} {u}^{2} (1 - { \sin^2 \alpha } + \frac{ { \sin^2 \alpha } }{2} ) \\ \\ = > 5(1 - { \sin^2 \alpha } ) = 2 - { \sin^2 \alpha } \\ \\ = > 3 = 4 { \sin^2 \alpha } \\ \\ = > \sin \alpha = \frac{ \sqrt{3} }{2} \\ \\ = > \alpha = \frac{ \pi}{3}$

$\bigstar\:\rm{\green{\boxed{\alpha\:=\:60^{\degree}\:}}}$

✍️ Therefore, the angle of projection $\rm{\alpha}$ with the horizontal is ‘60°’ .

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GIVEN :-

FORMULA :-

CALCULATION :-

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