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?

Question

Arun · Answer

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 · Answer

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

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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?

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?

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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?

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?

2 Answers

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