A high diver of mass 50.0 kg steps off a board 10.0 m above the water and falls vertical to the water, starting from rest. If her downward motion is stopped 2.10 s after her feet first touch the water, what average upward force did the water exert on her?
To find the average upward force exerted by the water on the high diver, we can use the equation of motion.
1. First, we need to calculate the time it takes for the high diver to fall from the board to the water using the equation of motion:
h = (1/2) * g * t^2
Where h is the height (10.0 m), g is the acceleration due to gravity (9.8 m/s^2), and t is the time.
Rearranging the equation to solve for t:
t = sqrt((2 * h) / g)
Substituting the values:
t = sqrt((2 * 10.0) / 9.8)
≈ 1.43 s
2. Since the downward motion is stopped 2.10 s after her feet first touch the water, the time during which the upward force acts is (2.10 - 1.43) = 0.67 s.
3. Now, we can calculate the average upward force using Newton's second law:
F = m * a
Where F is the force, m is the mass of the high diver (50.0 kg), and a is the average acceleration during the upward motion.
Since the high diver starts from rest and comes to a stop, the final velocity is zero. Therefore, the average acceleration can be calculated as follows:
a = (0 - 0) / (2.10 - 1.43)
≈ 0 m/s^2
Finally, substituting the values into the equation:
F = 50.0 kg * 0 m/s^2
= 0 N
Therefore, the average upward force exerted by the water on the high diver is 0 N.