The height in metres, of an object that has fallen from a height of 180m is given by the position function s(t) = -5t^2 + 180, when t greater than or equal to 0 and t is in seconds.
c) At what velocity will the object hit the ground?
How would you solve having previous knowledge of average rate of change, instantaneous rate of change, limits? I know how to get it using the kinematics equations, but I am having trouble doing this through pure math.
I know that d will equal 0 when this happens, and t I don't know. I solve for t with d being 0 and I get (t-6)(t+6), so t= 6s.
the veolocity at a time, t, is
Ok thank you, but I thought velocity is like this:
velocity = displacement / time
In this case if the object hit the ground, wouldn't the displacement be 0, and thus the numerator be 0? I am thinking this because when it hits the ground, shouldn't the given function = 0?