A small rock moves in water, and the force exerted on it by the water is given b

Question

A small rock moves in water, and the force exerted on it by the water is given by f=kv. the terminal speed of the rock is measured and found to be 2,0 m/s. The rock is projected upward at an initial speed of 6,0 m/s. You can ignore the buoyancy force on the rock.

a) In the absence of fluid reisistance, how high will the rock rise and how long will it take to reach this maximum height?

b) when the effects of fluid resistance are included, what are the answers to the questons in part(a)?

I managed a.

in b) I and the guided solution has the same solution for V(t) = V0e^(-k/mt)+Vt(1-e^(-k/mt))

But for Y the guided solution has Vt( t+m/k(e^(-k/mt)-1)) It lookes like the solution has not integrated V0e^(-k/mt).

My expression for t and the guided solution is the same: t =-m/k*ln((Vt-Vy)/(Vt-V0)), Vy= 0.

I take the absolute value of the expression inside ln. I think of Vt-V0 = absoute value of 2m/s-6m/s=-(-4m/s)=4m/s.

The guided solution: 2m/s-(-6m/s), but Vt and V0 has the same directon. M t = 0,14s and the guided solution t= 0,283.

My Y = 0,69m and the guided solution: Y = 0,69m.

Explanation / Answer

Vf^2 = Vo^2 + 2g*d,
d = (Vf^2 - Vo^2) / 2g,
d = (2^2 - 6^2) / -19.6 = 1.63m.

for parts a and b, you can use the same equations you had in first year physics to solve trajectories;

height = v0^2/2g = 1.78m

time to max height = v0/g = 0.6s

including friction, you have to solve the differential equation:

m dv/dt = - mg - kv where mg is the downward force of gravity and -kv the downard acting force of water friction

separate variables:

m dv/(mg + kv) = - dt

integrate both sides:

(m/k) ln[mg + kv] = - t +C where C is the constant of integration

multiply through and exponentiate both sides:

mg + kv = Aexp(-kt/m) where A is a constant

this gives us

v=1/k(Aexp(-kt/m) - mg)

we solve for k knowing that the terminal velocity is 2.5m/s; let t become very large so that

v=1/k*(-mg) =-2.5m/s so that k =mg/2.5

solve for A using v = 5.9 when t=0:

this will give you a complete expression for v(t); to find max height, recall that v(t) is the derivative of position, so you have max height when dy/dt =0; but we know that v = dy/dt so just find the time where v=0;

then integrate v(t) to find y(t); substitute this value of t into y(t) and compute the max height

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