Solving for v(t), tmax, and Max Height of a Thrown Ball

[zeromaxxx]

Homework Statement

A ball is thrown vertically upwards at speed v₀. Assume drag force is proportional to v².

a) Show that, while moving upwards, Newton's Second Law gives a = -g(1+v²/v_t²) where v_t is the terminal speed.

b) Take v₀ = 3v_t and solve for v(t), the time t_max at which it reaches max height and the maximum height it reaches. Express your results in terms of g and v_t

Homework Equations

a = -g(1+v²/v_t²)

v = -v_t tanh (t/T), where T = v_t/g

y = -v_t²/g * ln [cosh (gt/v_t)]

The Attempt at a Solution

I got part A of the question simple enough, I'm having trouble with part B. I got the expression for v(t) which was the integral of the first equation. I'm having trouble looking for t_max and the max height. It sounds simple enough but I can't grasp around the idea on how to do it.

 

[hotvette]

Hint: what is v at the max height? Think calculus.

 

[zeromaxxx]

What is v at the max height? Think calculus.

v at max height should be 0 right?

so then 0 = -v_t tanh (t/T)

It's how to deal with the equation that comes after.

tanh x = (e^x - e^-x)/(e^x + e^-x)

tanh (t/T) = (e^t/T - e^-t/T)/(e^t/T + e^-t/T)

0 = -v_t[ (e^t/T - e^-t/T)/(e^t/T + e^-t/T)]

so if v = 0, then the the exponential functions on the numerator should equal 1 since -v_t(1-1/1+1) = 0. This would be the case if the term (t/T) = 0 cause e⁰ = 1.

That's the farthest I got since I can't think of anything that can make the term equal zero unless ' t ' itself is zero. I think I'm missing something here.

 

[hotvette]

I believe your expression for v(t) is incorrect, since v(0) should be v₀.

 

[Dime]

I have the same question on my assignment. I don't get any hyperbolic functions because the object is moving upwards. It would become hyperbolic if it was moving downwards. The integral of 1/(1+x^2) is arctan. So instead of having tanh it would be changed to arctan.