Solving Problem on Motions 1 - Vf, Vi, t

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The discussion focuses on solving a physics problem involving a diver's motion from a 10m diving board. The calculations for part (a) reveal that the final velocity upon impact with the water is approximately 9.90 m/s, with an elapsed time of 1.01 seconds. The correct application of the kinematic equation Vf^2 = Vi^2 - 2g(yf - yi) is confirmed, although the final answer requires verification of arithmetic. Further parts of the problem involve integrating motion equations under water, considering buoyant and viscous forces.

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1. A guy with mass m, begins his descent from a diving board 10m high with zero initial velocity.

(a) Calculate his velocity on impact with the water, v0, and the elapsed time from his jump until impact. Assume there is no air resistance.

(b) Next under water, let’s suppose the buoyant force on the guy balances his gravitational force and he experiences a viscous force of bv^2. Solve for his velocity v in terms of the depth x under water, taking v=v0 at x=0. (Hint: You’re required to integrate the equation of motion for his vertical descent)

(c) Suppose b/m = 0.4/m, estimate the depth at which v = 0.1v0.

(d) Solve for his vertical depth x(t) in terms of the time under water.





(a) Vf^2 = Vi^2 - 2g (yf-yi)
V^2 = 0 - 2g (10m)
V = 9.90m/s
Vf = Vi - gt
t = 1.01s
I only know how to do (a) but I am not sure whether it is right. Please help!
 
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nashvision said:
(a) Vf^2 = Vi^2 - 2g (yf-yi)
V^2 = 0 - 2g (10m)
V = 9.90m/s
That's the correct formula, but your final answer is incorrect. Double check your arithmetic.
 

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