2. A 755 N diver drops from a board 10.0 m above the water\'s surface. a. Find t

Question

2. A 755 N diver drops from a board 10.0 m above the water's surface.
a. Find the diver's speed 5.00 m above the water's surface
b. Find the diver's speed just before striking the water.
3.If the diver in problem 2 leaves the board with an initial downward speed of 2.00 m/s, find the diver's speed when striking the water.
5. A pendulum 2.0 m long is released from rest when the support string is at an angle of 25.0 degrees with the vertical. What is the speed of the bob at the bottom of the swing?

Explanation / Answer

2. TE1 = TE2 PE1 + KE1 = PE2 + KE2 V1 = 0, h1 = 10.0 m, h2 = 10.0 - 5.00m = 5.0 m mgh1 + 0.5 mV1^2 = mgh2 + 0.5 mV2^2 gh1 + 0.5 V1^2 = gh2 + 0.5 V2^2 gh1 + 0 = gh2 + 0.5 V2^2 gh1 - gh2 = 0.5 V2^2 2g(h1 - gh2) = V2^2 V2 = sqrt(2g(h1-h2)) = sqrt(2*9.81*(10.0-5.00))m/s V3 = sqrt(2g(h1-h3)) = sqrt(2*9.81*(10.0-0.00))m/s 3. The speed gained from the acceleration due to gravity is v(2ax); add the initial speed (which does not change during the fall) to get 14.0 + 2.00 = 16.0 m/s. 5. Conservation of energy. Figure out the vertical distance between the first point and the bottom of the swing. mg(delta-h)=1/2*m*v^2 delta-h=2m-2m(cos25) delta-h=0.19m 9.81m/s^2*0.19m*2=v^2 v=1.93m/s

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