Question

A small rock moves in water, and the force on it by the water is given by f=kv. The terminal speed of the rock is measured and found to be 2.0m/s. The rock is projected upward at an inital speed of 6.0m/s. You can ignore the buoyancy force on the rock

a) in the absence of fluid resistance, how high will the rock rise and how long will it take to reach this maximum height? I managed this one. b) when the effects of fluidresistance are included, what are the answers to the question in part a).

i manage to get the same z time as the solution on this website.

I and the website has the same equation for Vy.

the websites solution for Y = Vt*[1-e^(-k/mt))] which is the same expression as the example in the book.

But in the example in the book there is no initial speed Vy=0. In the question there is initla speed and therefor I think the expression for Y must be different from the examples expression. Therefore I think the solution of this question on this website must be wrong. Please comment me on this.

Answer 1

for parts a and b, you can use the same equations

height = v0^2/2g = 2m

time to max height = v0/g = 0.61s

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 2m/s; let t become very large so that

v=1/k*(-mg) =-2m/s so that k =mg/2

solve for A using v = 6 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