Tuesday

March 3, 2015

March 3, 2015

Posted by **jake** on Wednesday, October 20, 2010 at 7:04pm.

- Physics -
**Henry**, Thursday, October 21, 2010 at 10:30pmThe 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^wrong!

**Answer this Question**

**Related Questions**

physics - A car drives straight off the edge of a cliff that is 58 m high. The ...

Physics - A car drives straight off the edge of a cliff that is 58 m high. The ...

Physics - A car drives straight off the edge of a cliff that is 58 m high. The ...

Physics - A car drives straight off the edge of a cliff that is 60 m high. The ...

Physics - A car drives straight off the edge of a cliff that is 68 m high. The ...

Physics - A car drives straight off the edge of a cliff that is 68 m high. The ...

physics - A car drives straight off the edge of a cliff that is 56 m high. The ...

Physics - A car drives straight off the edge of a cliff that is 50 m high. The ...

physics - A car drives straight off the edge of a cliff that is 47m hight. The ...

physics - a car is driven off the edge of that is 54.0 m high. the point of ...