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754 N diver drops from a board 9.00 m above the water's surface. The acceleration of...

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harris123 | Student, Undergraduate | (Level 2) eNoter

Posted November 16, 2009 at 9:31 PM via web

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754 N diver drops from a board 9.00 m above the water's surface.

The acceleration of gravity is 9.81 m/s^2.

Find the diver's speed 5.80 m above the water's surface. Answer in units of m/s.

Find the diver's speed just before striking the water. Answer in units of m/s.

If the diver leaves the board with an initial upward speed of 2.90 m/s, find the diver's speed when striking the water. Answer in units of m/s.

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neela | High School Teacher | (Level 3) Valedictorian

Posted November 16, 2009 at 11:19 PM (Answer #1)

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The speed v  (in mer per second) of a falling object from rest, given the height h in meter  can be calculated.

v^2 = u^2+2gh, wher u is the initial velocity V is the final velocity when a falling distance is h. g is the acceleration due to gravity.

A)

To find the speed when the diver is above 5.8m from water.

h = initial heiht of the diver - 5.8m = 4.8m, u=0

V^2 = U^2+2gh =0+2*9.81*4.8

v=sqrt(2*9.81*4.8) = 9.7044m/s

B)

The speed of the diver just before touching the water surface v = sqrt(2*9.81*9m), as here h=9 meter.

=13.2883 m/s.

C)

The initial speed is upward and so u = 2.9m/s as direction of moving towards water is assumed positive. This could be solved by using  equation of motion for the vertically upward projected . But the  magnitude of the speed remains same 2.9m/s  while crossing the board in downward direction.Threfre,

v^2=u^2+2gh, u=2.9m/s, g=9.81 and h = 9m

v = sqrt(2.9^2+2*9.81*9)

= 6.0852 m/s

 

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