Question

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

Answer 1

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

Answer 2

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