A 50 kg person steps off a diving platform that is 10 meters above the water below (Olympic height). With what speed do they hit the water?

14 m/s

Explanation:

The following data were obtained from the question:

Mass = 50 kg

Initial velocity (u) = 0 m/s

Height (h) = 10 m

Acceleration due to gravity (g) = 9.8 m/s²

Final velocity (v) =?

The velocity (v) with which the person hit the water can be obtained as shown below:

v² = u² + 2gh

v² = 0² + (2 × 9.8 × 10)

v² = 0 + 196

v² = 196

Take the square root of both side

v = √196

v = 14 m/s

Therefore, he will hit the water with a speed of 14 m/s

small = small current large mag large current