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?
1 answer:
Answer:
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
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