A car drives straight off the edge of a cliff that is 58 m high. The police at the scene of the accident note that the point of impact is 0.136 km from the base of the cliff. How far down had the car fallen after 1.5s?

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Answer 1 · 2015-09-16T17:04:12

#11"m"#

The car is falling under gravity with an initial vertical component of zero velocity so:

#s=1/2"g"t^2#

#s=1/2xx9.8xx1.5^2=11"m"#

The distance of the point of impact is irrelevant.