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Physics-

General

Easy

Question

From an inclined plane two particles are projected with same speed at same angle $theta$ , one up and other down the plane as shown in figure, which of the following statements is/are correct?

The time of flight of each particle is the same
The particles will collide the plane with same speed
Both the particles strike the plane perpendicularly
The particles will collide in mid air if projected simultaneously and time of flight of each particle is less than of collision

The correct answer is: The time of flight of each particle is the same

$H e r e comma blank alpha equals 2 theta comma blank beta equals theta$

$T i m e blank o f blank f l i g h t blank o f blank A blank i s comma$
$T subscript 1 end subscript equals fraction numerator 2 u sin invisible function application open parentheses alpha minus beta close parentheses over denominator g cos invisible function application beta end fraction$
$equals fraction numerator 2 u sin invisible function application open parentheses 2 theta minus theta close parentheses over denominator g cos invisible function application theta end fraction$
$equals fraction numerator 2 u over denominator g end fraction tan invisible function application theta$
$T i m e blank o f blank f l i g h t blank o f blank B blank i s comma blank T subscript 2 end subscript equals fraction numerator 2 u sin invisible function application theta over denominator g cos invisible function application theta end fraction$
$equals fraction numerator 2 u over denominator g end fraction tan invisible function application theta$
So, $T subscript 1 end subscript equals T subscript 2 end subscript$ . The acceleration of both the particles is $g$ downwards. Therefore, relative acceleration between the two I zero or relative motion between the two is uniform. The relative velocity of $A$ w.r.t. $B$ is towards $A B$ , therefore collision will take place between the two in mid air.

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