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?
Manjeet , 5 Years ago
Grade 12th pass
2 Answers
Arun
Last Activity: 5 Years ago
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
Manjeet
Last Activity: 5 Years ago
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
Provide a better Answer & Earn Cool Goodies
Enter text here...
LIVE ONLINE CLASSES
Prepraring for the competition made easy just by live online class.
Full Live Access
Study Material
Live Doubts Solving
Daily Class Assignments
Ask a Doubt
Get your questions answered by the expert for free