Need help with solving this physics problem

Question

A large rock is dropped from the top of a high cliff. Assuming that air resistance can be ignored and that the acceleration has the constant value of 10 m/s2, how fast would the rock travel in 6 seconds after it is dropped? What is this speed in MPH?

Answer 1

s(t) = -(10/2)t^2

in 6 seconds

s(6) = -5(36) = -180 meters per 6 second

= -180/6 = average speed of -30 m/s = about -67 mph

10m/s^2 is an approximation. 9.8 m/s^2 is the more common and more accurate approximation for the effect of gravity

that would give -4.9(6^2) = -4.9(36) = -176.4 meters in 6 seconds -176.4 m/6 s = average -29.4 meters per second

negative sign just means downward direction

speed at the instant in time of 6 seconds = derivative of s(t) = v(t) = 10t = 10(6) = 60 meters per second, twice the average speed