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