A partical is projected with a certain angle "Alpha" above the horizontal from t...

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September 2021 1 23 Report

A partical is projected with a certain angle "Alpha" above the horizontal from the foot of an inclined plane of inclination 30⁰. If the partical strikes the plane normally, find "Alpha"

Topic - Kinematics

Answers & Comments

SparklingBoy

⇝ Given :-

A partical is projected with a certain angle "Alpha" above the horizontal from the foot of an inclined plane of inclination 30⁰.

The partical strikes the plane normally.

⇝ To Find :-

Value of Alpha.

⇝ Solution :-

Let the angle of projection with respect to in client plane be = θ.

Therefore,

$\: \: \: \: \: \: \purple{ \large \underline {\boxed{{\pmb{ \theta = ( \alpha - 30 \degree)}} }}} \\$

Here Tine of Flight of particle is :

$\rm \: T = \frac{2u \sin \theta}{g \cos \theta} \\$

⏩ Putting Value of θ :

$\red{\bf T = \frac{2usin ( \pmb\alpha - 30 \degree)}{gcos30 \degree}} \: - - -(1) \\$

Equation of motion in x - direction along plane is ;

$\large\rm v_x=u_x+a_xT$

As the particle strikes the plane Normally

Therefore at point B velocity in x - direction should be Zero.

$:\longmapsto \rm 0 = u \: cos \theta + ( - g \: sin30 \degree)T \\$

$:\longmapsto \rm u \: cos( \alpha - 30 \degree) - g \: sin30 \degree T = 0 \\$

⏩ Putting Value of T from (1) ;

$:\longmapsto \rm ucos( \alpha - 30 \degree) - \frac{ \cancel{g}}{ \cancel2} \times \frac{ \cancel2usin( \alpha - 30 \degree)}{ \cancel{g}cos30 \degree} = 0 \\$

$:\longmapsto \rm ucos( \alpha - 30 \degree) = \frac{u[sin( \alpha - 30 \degree)]}{cos30 \degree} \\$

$:\longmapsto \rm tan( \alpha - 30 \degree) = \frac{ \sqrt{3} }{2} \\$

$:\longmapsto \rm \alpha - 30 \degree = \tan {}^{ - 1} \bigg( \frac{ \sqrt{3} }{2} \bigg) \\$

$\large\underline{\pink{\underline{\frak{\pmb{\alpha = {tan}^{ - 1} \bigg( \frac{ \sqrt{3} }{2} \bigg) + 30 \degree }}}}} \\$

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