What is the velocity of an object dropped from a height of 180 m when it hits the ground?

V^2 = initial velocity^2 + 2 * acceleration * distance

v^2 = (0m/s) + 2 * 9.8 m/sec^2 * 180 m

v^2 = 3528 m^2 / sec^2

v = 59.397 m/sec

You can use this formula: vf^2=0+2 (9.8) (M)

solve it out like so:

vf^2=0+2 (9.8) (180)

vf^2=2 (1764)

vf^2=3528

now, square both sides to cancel out the power of 2. you are left with:

about 59.4 m/s