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It is possible to project a particle with a given velocity in two possible ways so as to make it pass through a point P at a distance r from the point of projection. The product of the times taken to reach this point in the two possible ways is then proprtional to? 1/r r r^3 1/(r)^2

It is possible to project a particle with a given velocity in two possible ways so as  to make it pass through a point P at a distance r from the point of projection. The product of the times taken to reach this point in the two possible ways is then proprtional to?


1/r


r


r^3


1/(r)^2

Grade:

2 Answers

Rohan Das
36 Points
8 years ago

WE KNOW THAT RANGE OF A PROJECTILE FOR ANY  ANGLE EQUALS 180DEGREE MINUS TWO TIMES THE ANGLE.

SO TIME DEPENDS UPON ANGLE OF PROJECTION THAT IS THETA AND ANGLE DEPENDS UPON R.

SO TIME IS DIRECTLY PROPORTIONAL TO r.

Akash Kumar Dutta
98 Points
8 years ago

Dear Shayan,
projectiles are fire at (x) and (90-x) angles to reach the same horizontal range.
t1=r/ucosx
t2=r/ucos(90-x)=r/usinx
t1.t2= r^2/u^2sinxcosx=2r^2/u^2 sin2x
multipling g up and down, and putting r=u^2sin2x/2g we get....
t1.t2= g(r^2) / r=gr
hence t1.t2 are proportional to r (ANS)
Regards.

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