Question

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​

Answer 1

⇝ 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) ;

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

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

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

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

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

Answer 2

Given :-

• A particle is projected with a certain angle 'Alpha ' above the horizontal from the foot of an inclined plane of inclination of 30 degree.If the partical strikes the plane normally .

□To find :-

• Here we should the value of Alpha

□Solution :-

• Here u can refer the attachment for more information.
• It is better to understand.

□Used formula :-

• $T = \frac{2u \: sin \: theta}{g \: cos \: theta}$
• Here we used the formula to get the solution of answer.

□Hope it helps u ,@splendidcharm.

□Thank you