Question

A high diver of mass 60.0 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

Answer 1

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;

$F_{on\ diver} = mg - F_{ \ water}\\\\ma = mg - F_{ \ water}\\\\F_{ \ water} = mg - ma\\\\F_{ \ water} = m(g-a)\\\\F_{ \ water} = 60[9.8-(-6.67)]\\\\F_{ \ water} = 60 (9.8+6.67)\\\\F_{ \ water} = 60(16.47)\\\\F_{ \ water} = 988.2 \ N$

Therefore, the average upward force exerted by the water is 988.2 N