Question

A body is protected vertically upward with speed 40 m /s.the distance travelled by body in the last second of upward journey.??

Answer 1

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HERE IS YOUR ANSWER,


=> Distance travelled by body in last second of “upward” journey = Distance travelled by body in first second of “downward” journey.

=> Distance travelled by body in first second of “downward” journey

S = 0.5at^2

=> 0.5 * 9.8 * 1^2

=> 4.9 m

∴ Distance travelled by body in last second of upward journey is 4.9 m (or 5 m if g = 10 m/s^2)


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