which yields F≈2.8 ×104kg. Since we are not at all certain of his mass, we express this as a guessed-at range (in kN) 25 < F< 30. SinceFmg>>, the impulse GJdue to the net force (while he is in contact with the water) is overwhelmingly caused by the upward force of the water: FdtJ=zGto a good approximation. Thus, by Eq. 9-29, Fdtppmghfi=−=−−zGG02di(the minus sign with the initial velocity is due to the fact that downward is the negative direction) which yields ()23(70 kg) 2 9.8 m/s12 m1.1 10 kg m s.=×⋅Expressing this as a range we estimate 331.0 10 kg m s1.2 10 kg m s.×⋅<<³25. We estimate his mass in the neighborhood of 70 kg and compute the upward force Fof the water from Newton’s second law:
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This note was uploaded on 05/17/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.