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Grade: 12th pass
        
A particle is projected vertically upward from A with a speed of 50m/s and another is dropped simultaneously from B , which is 200m vertically above A. They will cross each other after what time?
5 months ago

Answers : (2)

Arun
20881 Points
							
Let them meet at a point h above the ground at a time t. 
• Distance moved down by B in t s = (0.5gt^2) .= 200 - h 
Distance moved up by A in t second is 50 t – 0.5gt^2.= h 
200 = 50 t 
• t = 4 s 
5 months ago
Manjeet
14 Points
							
Let h be the distance from A at which the two particles meet and t be the time taken hence that for B will be 200-h
For particle projected upward equation of motion can be given as:
h=50t+0.5gt^2
For B 
-0.5gt^2 =200-h
Solving above equations
h =50t-(200-h)
0=50t-200
t=4s
 
5 months ago
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