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The vertical distance covered by a free falling object is given by the formula: S= ut+0.5at^2, where S is the distance covered (height of the building), u is the initial velocity (for this case it is 0 since the body is released from rest), t is the time taken for the object to hit the ground (it has taken 5 seconds) and a is the acceleration due to gravitational pull (assumed to be 9.8ms^2). Therefore, the height of the building is given by (0x5 +0.5x9.8 x25) which is 122.5m.

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9y ago
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14y ago

x = 1/2 G t2 = (1/2) (32.2) (9) = 144.9 ft (44.2 meters) (rounded)

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7y ago

54 m/s

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Q: What if a ball is dropped off the roof of a tall building if the ball reaches the ground in 3 seconds how tall is the building?
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