A diver whose gender is not relevant jumps upward at 4 m/s from a 10 m diving board. How fast is he/she moving when he/she strikes the water? Answer in m/s.

The final velocity of the diver can be found using the formula:

vf^2 = vi^2 + 2ad

where vf is the final velocity, vi is the initial velocity (4 m/s), a is the acceleration due to gravity (-9.8 m/s^2), and d is the distance the diver falls (10 m).

Plugging in the values, we get:

vf^2 = (4 m/s)^2 + 2(-9.8 m/s^2)(10 m)

vf^2 = 16 m^2/s^2 - 196 m^2/s^2

vf^2 = -180 m^2/s^2

(note that the negative sign just indicates the direction of motion, not the magnitude)

Since the final velocity cannot be negative, we know that the diver does not actually reach the water at this speed. Instead, we can use the absolute value of the final velocity:

|vf| = sqrt(180 m^2/s^2)

|vf| = 13.42 m/s

Therefore, the diver is moving at approximately 13.42 m/s when he/she strikes the water.