A construction worker working at 300 feet from the ground, drops a nail onto the ground below. The function, s(t) = -4.9t2 + 300, gives the height (in meters) of the object as it falls to the ground. The velocity at time t = 3 seconds is given by lim(t->3) s(3)- s(t)/ 3-t. Evaluate this limit to find the velocity of the nail at 3 seconds.

To evaluate the given limit, we need to find the velocity of the nail at 3 seconds, which is represented by the derivative of the height function, s(t).

First, let's find the derivative of s(t) with respect to t.
The derivative of -4.9t^2 is -9.8t.
The derivative of 300 is 0.

So, the derivative of s(t) is s'(t) = -9.8t.

Now, let's substitute 3 into the derivative function to find the velocity at t = 3 seconds.
s'(3) = -9.8 * 3 = -29.4.

Therefore, the velocity of the nail at 3 seconds is -29.4 meters per second.