3hey siddhart is this ur dream question.
Rajnikanth faces 2 villians opposite to him. He has one bullet( diagram given in box). He throws a coin with a velocity 3 m/s and accelaration 2 m/s2. He shoots a bullet with 200 m/s and this bullets hits the coin in 2 seconds. The bullets breaks into 2 halfs. One with electron and the other with a proton. The one with electron has 100 m/s velocity. A magnetic feild of 10-10 T acts into the screen or -k direction. How far away are villian 1 and villian 2 from rajnikanth if the bullets hit them just before the bullets form a quarter of a circle? neglect gravity. Usage of calculator is allowed.
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14 Answers
9by quantum thry u cant predict the pos of villians cos unpredictability factor for Rajnikanth medium ≈1
11It doesn't matter where the villains are. Rajnikanth's magic bullets will always hit them.
So cool down and leave the physics part of it to Rajnikanth. [4]
62haha.. this is a question :D
And what a comment by celestine :D
by quantu thry u cant predict the pos of villians cos unpredictability factor for Rajnikanth medium ≈1
11Well, Siddharth. It was your idea to make this question fun. Look now. Everyone's ROFL instead of solving the question [9]
1Well u have a little bit of kinematics too.. I wanted to use projectile motion but decided not too because it would become too complex.Anyway its Rajnikanth.. nothing is impossible for him..
MORE QUESTIONS TO COME ON OUR DEAR RAJNIKANTH...
11Do you need horizontal distance or actual distance?
Hope the question doesn't sound silly cos finding the "horizontal distance" is a whole lot easier. [4]
11We can also find the "actual distance" if you want to. It's only the hypotenuse of the Δle with other sides as horizontal distance and the radius.[1]
Horizontal Distance= Distance travelled by the bullet + radius of the particle's path
I'm too lazy to do the calculations [3] so all the best [4]
1Any way its the right way to do aragorn... Great.. Keep it up bro...