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 from the cliff was the car after 1.5s?

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Answer 1 · 2015-11-05T21:55:10

$59.3 m$ $59.3"m"$

We can find the time of flight by considering the vertical component of the motion:

$s = \frac{1}{2} g t^{2}$ $s=1/2"g"t^2$

$t^{2} = \frac{2 s}{g}$ $t^2=(2s)/"g"$

$t = \sqrt{\frac{2 \times 58}{9.8}}$ $t=sqrt((2xx58)/(9.8))$

$t = 3.44 s$ $t=3.44"s"$

The horizontal component of velocity is constant so:

$v = \frac{s}{t} = \frac{136}{3.44} = 39.53 m/s$ $v=s/t=136/3.44=39.53"m/s"$

So after 1.5s we get:

$s = v \times t = 39.53 \times 1.5 = 59.3 m$ $s=vxxt=39.53xx1.5=59.3"m"$