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 m

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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;

   [tex]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[/tex]

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

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