A.) 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.
B.) Find the diver's speed 5.80 m above the water's surface. Answer in units of m/s.
"A 754N diver drops from a board 9.00 m above the water's surface. The acceleration of gravity is 9.81 m/s^2."
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The motion of the diver upward with a initial velocity of 2.9m/s.
Taking the motion directed upward as postive, the equation of mtion of the diver with initial speed u, and displecemnt s intime t is given by
s=ut-(1/2)gt^2., wher g is the gravitational acceleration and is equal to 9.81m/s^2 and u =2.9.
When the diver reches the water below 9m , s= -9m So the equation becomes:
-9 = 2.9*t-0.5*9.81t^2 0r
4.905t^2-2.9t-1.3864539 = 0. We solve for t, time taken by the diver to reach the water.
t =[ 2.9 + sqrt(2.9^2+4*4.905*9)]/(2*4.905)
Therefore, the speed when he strike water = u-gt =2.9-9.81(1.68207)= -13.6011m/s
The divers speed at 5.5 m above the water surface . So s = -3.5 m below the board.
-3.5= 2.9t- 4.905^2 or
t = 1.19057 s
v=2.9-9.81(1.19057) = -8.7795 m/s
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