Question

A diver drops from a 10 meter high cliff. At what speed does he enter the
water, and how long is he in the air?

Answer 1

\[0.5(-10m/s^2)t^2 + 0m/s(t) + 10m = 0\]
Here are all three factors - acceleration (gravity), velocity (there is no velocity), and height (10 meters). If we set all that equal to zero, then we get the time that diver takes to reach a height of 0m (the water level).

\[0.5(10m/s^2)t^2 = 10m\]
\[t = \sqrt{2}s\]

To find the speed, we substitute this time into gravity...
\[-10m/s^2(\sqrt{2}s) = -10\sqrt{2}m/s \approx -14.14m/s\]