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Two particles are projected from two towers simulta as shown what should be the value of d for their collision

Two particles are projected from two towers simulta as shown what should be the value of d for their collision 

Question Image
Grade:11

2 Answers

Arun
25763 Points
3 years ago
d = 20 m, for every value of time the collision takes 

You can make a drawing to simplify the problem, but this is simple: 

The difference of height between the towers is of 10 m, and let's say the collision takes 1 second, the particle of tower 20m goes at 10m/s, with that you have that one part of d is 10 m 

Now, you need to find the other part with trigonometry, you have the triangle's height that is 10, and the hypotenuse that is 14.14, and you have the angle of 45 so to find the base you do: 
tan 45 = 10/x x=10 

The other horizontal part is 10, you add it to the first one that was also 10, so the towers are separated 20 m each other
 
 
Sarthak Jain
26 Points
14 days ago
Use the concept of relative motion
Let the distance be d 
Component of velocity A w.r.t. B is 20i +10j
Required height = final - initial = 10 
Tan theta =10/d =20/10
10d = 200
D=20 m 

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