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three particles A ,b and C are thrown from the top of a tower with the same speed .A is thrown straight up.B isn thrown straight down and C is thrown horizontly they hit the ground with speed v A , V B ,and v c then compare the three velocities

three particles A ,b and C are thrown from the top of a tower with the same speed .A is thrown straight up.B isn thrown straight down and C is thrown horizontly they hit the ground with speed vA ,VB ,and vc then compare the three velocities 

Grade:12

3 Answers

SAHIL
3778 Points
7 years ago
When Particle A arrives back at the top of the tower, it will have the same downward speed as the upward speed it started with; its subsequent motion will be identical to that of Particle B. That va and vb are > vc is clear from the fact that A is freely "dropped" from its maximum height, which is greater than the height of the tower, hence greater than max height of C.
SO ANSWER IS
va = vb > vc
THANKS
Guru
9 Points
7 years ago
For A , Va = u + gt
 
For B, Vb = u + gt
 
For c, Vx = u ( g= 0 )
          Vy = gt ( Uy = 0 )
 
For  Vc = root Vx^2 + Vy^2
               = root u^2  + gt ^2
               = u + gt 
Hence  Va = Vb = Vc. 
Rajat
11 Points
6 years ago
Va^2 =U^2 +2(-g)(-h)=U^2 +2ghVb^2 =U^2 +2ghFor CVc in x direction=UVc in y direction=√2ghVc^2= (Vc)x^2 + (Vc)y^2 =U^2 + 2ghVa^2=Vb^2= Vc^2Va=Vb= Vc

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