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A particle is projected at an angle 60 degree,with speed 10sqrt3m/s from the Point A towards right.At the same time the wedge (HAVING ITS FOOT of inclination 30 degree at A) is made to move with speed 10sqrt3m/s towards right.Then the time after which the particle will strike with the wedge is (a)2s (b2sqrt3 m/s (c)4/sqrt3 m/s (d)none .............

A particle is projected at an angle 60 degree,with speed 10sqrt3m/s from the Point A towards right.At the same time the wedge (HAVING ITS FOOT of inclination 30 degree at A) is made to move with speed 10sqrt3m/s towards right.Then the time after which the particle will strike with the wedge is


(a)2s


(b2sqrt3 m/s


(c)4/sqrt3 m/s


(d)none .............

Grade:12

3 Answers

AKASH GOYAL AskiitiansExpert-IITD
419 Points
11 years ago

dear Rajan

resolve the velocity of projectile

Ux= 10√3 COS60= 5√3

Uy=10√3 sin60=15

now you can see that horizontal velocity of wedge is more than the projectile hence they will not meet.

answer will be (d) none

All the best.

AKASH GOYAL

AskiitiansExpert-IITD

 

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apoorv
19 Points
5 years ago
wrong yr ye solution galat h iit exper bhai 
relative se ban gaya gadhe 
the answer is 2
1000
 

Please feel free to post as many doubts on our discussion forum as you can. We are all IITians and here to help you in your IIT JEE preparation.

 

Please feel free to post as many doubts on our discussion forum as you can. We are all IITians and here to help you in your IIT JEE preparation.

 
Shubham Maurya
23 Points
4 years ago
Relative se solve kar, ban jayega.Wedge ko rest mein la,Wedge ki velocity projectile ke horizontal component ko opposite direction mein de, fir solve kr ...and ans is 2 sec

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