Answer:
The average upward force exerted by the water is 988.2 N
Explanation:
Given;
mass of the diver, m = 60 kg
height of the board above the water, h = 10 m
time when her feet touched the water, t = 2.10 s
The final velocity of the diver, when she is under the influence of acceleration of free fall.
V² = U² + 2gh
where;
V is the final velocity
U is the initial velocity = 0
g is acceleration due gravity
h is the height of fall
V² = U² + 2gh
V² = 0 + 2 x 9.8 x 10
V² = 196
V = √196
V = 14 m/s
Acceleration of the diver during 2.10 s before her feet touched the water.
14 m/s is her initial velocity at this sage,
her final velocity at this stage is zero (0)
V = U + at
0 = 14 + 2.1(a)
2.1a = -14
a = -14 / 2.1
a = -6.67 m/s²
The average upward force exerted by the water;
![F_{on\ diver} = mg - F_{ \ water}\\\\ma = mg - F_{ \ water}\\\\F_{ \ water} = mg - ma\\\\F_{ \ water} = m(g-a)\\\\F_{ \ water} = 60[9.8-(-6.67)]\\\\F_{ \ water} = 60 (9.8+6.67)\\\\F_{ \ water} = 60(16.47)\\\\F_{ \ water} = 988.2 \ N](https://tex.z-dn.net/?f=F_%7Bon%5C%20diver%7D%20%3D%20mg%20-%20F_%7B%20%5C%20water%7D%5C%5C%5C%5Cma%20%3D%20mg%20-%20F_%7B%20%5C%20water%7D%5C%5C%5C%5CF_%7B%20%5C%20water%7D%20%3D%20mg%20-%20ma%5C%5C%5C%5CF_%7B%20%5C%20water%7D%20%3D%20m%28g-a%29%5C%5C%5C%5CF_%7B%20%5C%20water%7D%20%3D%2060%5B9.8-%28-6.67%29%5D%5C%5C%5C%5CF_%7B%20%5C%20water%7D%20%3D%2060%20%289.8%2B6.67%29%5C%5C%5C%5CF_%7B%20%5C%20water%7D%20%3D%2060%2816.47%29%5C%5C%5C%5CF_%7B%20%5C%20water%7D%20%3D%20988.2%20%5C%20N)
Therefore, the average upward force exerted by the water is 988.2 N
