Asked by jake

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 126 m from the base of the cliff. How fast was the car traveling when it went over the cliff?
Answered by Henry
The height of the bldg and the point
where the car hits the ground form the ver and hor sides of a rt triangle.
The distance is = to the hyp.

tanA = y / x,
tanA = 58 / 126 = 0.4603,
A =24.7 deg,

d = x /cosA = 126 / cos24.7 =138.7 m.

V^2 = 2gd,
V^2 = 2 * 9.8 * 138.7 = 2718.5,
V = sqrt(2718.5) = 52.1 m/s.
Answered by Anonymous
^wrong!
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