Answer:
851.33 N
Explanation:
Using newton;s equation of motion,
v² = u² + 2gh ......... Equation 1
Where v = final velocity, u = initial velocity, g = acceleration due to gravity, h = height
Given: u = 0 m/s(from rest), h = 10 m g = 9.8 m/s².
Substitute into equation 1
v² = 0² + 2(9.8)(10)
v² = 196
v = √196
v = 14 m/s.
Note: As the Diver touches the water, u = 14 m/s and v = 0 m/s( stopped)
Using,
d = (v+u)t/2 .............. equation 2
Where d = distance moved by the diver in water before its motion stopped, t = time taken before it comes to rest
Given: v = 0 m/s, u = 14 m/s t = 2.10 s
Substitute into equation 2
d = (0+14)2.1/2
d = 14.7 m.
Finally
work done by the water to stop the diver = potential energy of the diver
F×d = mgh'................Equation 3
Where F = force of the diver in water, d = distance of the diver in the water, m = mass of the diver, g = acceleration due to gravity, h' = height of the diver from the point of fall to the point where he comes to rest
making F the subject of the equation,
F = mgh/d ............ Equation 4
Given: m = 51.7 kg, h = 10 m, g = 9.8 m/s², d = 14.7 m.
Substitute into equation 4
F = 51.7(10+14.7)(9.8)/14.7
F = 851.33 N
Hence the upward force the water exert on her = 851.33 N
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