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A person jumps off a diving board 5.0 m above the water's surface into a deep pool. The person's downward motion stops 2.2 m below the surface of the water. Calculate the speed of the diver just before striking the water, in m/s (ignore air friction).

User Arocks
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2 Answers

4 votes

Final answer:

The speed of the diver just before striking the water is 11.88 m/s.

Step-by-step explanation:

To calculate the speed of the diver just before striking the water, we can use the concept of conservation of energy. At the top of the dive, the diver has potential energy, which is converted to kinetic energy as the diver falls. When the diver reaches the surface of the water, all of the potential energy has been converted to kinetic energy. We can use the equations for potential and kinetic energy to solve for the speed of the diver.

Given the height of the diving board (5.0 m) and the depth the diver stops (2.2 m below the surface), we can calculate the total distance the diver falls (5.0 m + 2.2 m = 7.2 m). Using the equations for potential energy (PE = mgh) and kinetic energy (KE = 0.5mv^2), we can set the two equal to each other and solve for the speed (v) of the diver:

PE = KE

mgh = 0.5mv^2

mg(7.2 m) = 0.5mv^2

gh = 0.5v^2

v^2 = 2gh

v = sqrt(2gh)

Substituting the values for g (acceleration due to gravity) and h (height of fall), we can calculate the speed of the diver:

v = sqrt(2 * 9.8 m/s^2 * 7.2 m) = sqrt(141.12) m/s = 11.88 m/s

User Aycan
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6.9k points
5 votes

Answer:

9.9 m/s

Step-by-step explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.81 m/s²


v^2-u^2=2as\\\Rightarrow v=√(2as+u^2)\\\Rightarrow v=√(2* 9.81* 5+0^2)\\\Rightarrow v=9.9\ m/s

If the body has started from rest then the initial velocity is 0. In order to find the velocity just before hitting the water then the distance at which the downward motion stops is irrelevant.

Hence, the speed of the diver just before striking the water is 9.9 m/s

User Any
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