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?
Physics - Henry, Thursday, October 21, 2010 at 10:30pm
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.
Physics - Anonymous, Sunday, February 5, 2012 at 6:58pm