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# A ball rolls off a 6m ledge and lands 4m away from the edge of the ledge. What was the ball's initial velocity?

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- NCSLv 74 weeks agoFavourite answer
vertically we can use

h = ½gt²

so time to fall t = √(2h/g) = √(2*6m / 9.8m/s²) = 1.1 s

horizontal velocity was

v = d / t = 4m / 1.1s = 3.6 m/s

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- frombrumLv 74 weeks ago
roughly

the ball falls at 9.81m per second

so 6/9.81 = .61 seconds of flight

the ball must have travelled 4m/.61 seconds is the velocity

resolve to m/s for the correct unit

(.61 is not accurate - use the equations of motion to get the accurate number)

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