A diver of mass 60kg, on entering the water with an initial velocity of 14.87ms-

Question

A diver of mass 60kg, on entering the water with an initial velocity of 14.87ms-1, experienced a force of resistance due to the opposing motion of magnitude of 180v Newtons, where v is the velocity of the diver in ms-1 . In addition to the force of gravity, the diver also experiences a constant upward buoyancy force of 720N. Take the origin of motion to be the surface of the water. The magnitude g of the acceleration due to gravity can be taken as 10ms-1 . Calculate the maximum depth, in metres the diver will submerge. Give answer to 3 decimal places.

Explanation / Answer

Let the required Distance be s

Resistance due to the opposing motion =- ma = 180v

Therefore

60*-a = 180v

-ve sign due to deceleration

Therefore

a = -3v

dv/ds =- 3v

dv/v = -3ds

Integrate both sides

lnv = -3s + C

As at s = 0 , v = 14.87 therefore

C = 2.7

Therefore

lnv = -3s + C

v = e2.7-3s

At it stops Vertical forces becomes in Equilibrium

Therefore

mg - 180v - 720 = 0

Therefore

600 - 180*e2.7-3s - 720 = 0

180e2.7-3s = - 120

2.7 - 3s = - 0.666

Therefore

s = 1.122 m

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