A swimmer runs over a horizontal diving board at a speed of
0.95 meters/second. 2.0 seconds after she dives, she hits the water. What was the height of the diving board from the water's surface and how far from the board was the swimmer when she hit the water?

Question

A swimmer runs over a horizontal diving board at a speed of
0.95 meters/second.  2.0 seconds after she dives, she hits the water. What was the height of the diving board from the water's surface and how far from the board was the swimmer when she hit the water?

festinger · Answer

$$s_{y}=u_{y}t+\frac{1}{2}at^{2}$$
Assuming uy is 0, putting values t=2, a=9.8, height is 19.6m.

Since she drops off at 0.95m/s horizontally, and stays midair for 2s, when she reaches the water she will be 0.95*2=1.9m away from where the board ends.