A ball is dropped from a high rise platform at t=0 starting at rest.After 6 seconds another ball is thrown downwards from the same platform with speed v.The 2 balls meet at t=18s.What is the value of v?

Question

Arun · Answer

Distance of 1st ball= S= 1/2 x 10 x 18 = 1620 Distance of 2nd ball = S' = ut + 1/2gt² 1620 = 12u + 5 x 144 u = 135 - 60 = 75 m/s

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A ball is dropped from a high rise platform at t=0 starting at rest.After 6 seconds another ball is thrown downwards from the same platform with speed v.The 2 balls meet at t=18s.What is the value of v?

A ball is dropped from a high rise platform at t=0 starting at rest.After 6 seconds another ball is thrown downwards from the same platform with speed v.The 2 balls meet at t=18s.What is the value of v?

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A ball is dropped from a high rise platform at t=0 starting at rest.After 6 seconds another ball is thrown downwards from the same platform with speed v.The 2 balls meet at t=18s.What is the value of v?

A ball is dropped from a high rise platform at t=0 starting at rest.After 6 seconds another ball is thrown downwards from the same platform with speed v.The 2 balls meet at t=18s.What is the value of v?

1 Answers

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Other Related Questions on Mechanics

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