Question

what is the velocity of a rock dropped from a height of 310m when it hits the ground?

Answer 1

v_{0}=0, x_{0}=310, a_{g}=-9.8, x=0
Therefore, -310=-(1/2)(9.8)t^2, and solve for t. Plug into v=v_{0}+a_{g}t. Profit!

Answer 2

Acceleration is 9.8 m/s^2. x(t) = x(0) + v(0)t +(1/2)at^2.
So, here,
x(t) = 310 + 0 + (1/2)(-9.8)t^2
This means 4.9t^2 = 310.
So t^2 = 310/4.9.
So t = 7.95 seconds is the time it hits the ground.
This means the velocity is 77.95 m/s at impact. (Time and veloc. rounded to nearest hundredth.)