Question

A 755 N diver drops from a board 10.0 m above the water’s surface. Find
the diver’s speed 5.00 m above the water’s surface. Then find the diver’s
speed just before striking the water.

Answer 1

F = 755N, and h = 10.0 m and a = 9.8 m/s^2 and m =? and v = ?To find the mass use F = ma 755 = (m)(9.8)  and m = .77 kg then use the formula KE + PE = PE2 + KE2 which is 1/2mv2 + mgh = mgh(2) + 1/2mv2 (2) to solve for the velocity. v1 = 0 and h1 = 10 and v2 = ? and h2 = 0 (1/2)(.77)(0^2) + (.77)(9.8.)(10) = (1/2)(.77)(V2^2) + (.77)(9.8)(0)v=14 m/s

Answer 2

We need to find the velocity of the diver at 5 m above the water and at the water surface.

Diver's velocity at 5 m from the surface is 9.9 m/s and at the surface is 14 m/s.

u = Initial velocity = 0

v = Final velocity

a = g = Acceleration due to gravity = $9.81\ \text{m/s}^2$

t = Time taken

s = Displacement

At 5 m from the surface s = 5

$v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 9.81\times5+0}\\\Rightarrow v=9.9\ \text{m/s}$

Diver's velocity at 5 m from the surface is 9.9 m/s.

At the surface s = 10

$v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 9.81\times10+0}\\\Rightarrow v=14\ \text{m/s}$

Diver's velocity at the surface is 14 m/s.

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