Kinematic Question: Finding Time to Reach Max Height with k, g, u

[gaobo9109]

Homework Statement

A particle of mass m is projected vertically upwards, with an initial speedo of u, in a medium which exerts a resistance of magnitude mkv, where k is a positive constant and v is the speed of the particle after time t. Express, in terms of k, g and u, the time taken for the particle to reach its greatest height

Homework Equations

u - gt = 0

The Attempt at a Solution

u - gt - kvt = 0
I don't know how to express v in terms of u, g and k.

 

[kuruman]

Can you find an expression for the acceleration of the particle?

 

[gaobo9109]

a = g + kv?

 

[kuruman]

Basically yes, but I would put a negative sign in front of both terms. Now can you write an expression for the rate of change of velocity with respect to time?

 

[gaobo9109]

Isn't dv/dt = a = -g - kv?

 

[kuruman]

Isn't dv/dt = a = -g - kv?

Correct. Now can you solve this equation to find v(t)?

 

[gaobo9109]

dv/dt = -g - kv
dv = -g - kvdt
Integrating both side
v = -gt - kvt
v = -gt / (1 + kt)

I just learned about differential equation, I am not sure if this is correct

 

[kuruman]

It is not correct. You need to separate the two variables before you integrate. This means that you need to rearrange the equation algebraically so that the left side has only v (and constants) in it and the right side has only t (and constants) in it.

 

[gaobo9109]

Sorry, I am quite new in differential equation. I don't know how to rearrange the equation such that v is on one side and t on the other side. Can you please teach me? Thanks

 

[kuruman]

This part of the process is "algebra" not "differential equations".

\frac{dv}{dt}=-(g+kv)

Cross multiply to get

\frac{dv}{g+kv}=-dt


Voila. The variables have been separated. Now you can integrate. Differential equations are not as scary as they sound. Now you can integrate.