Comparison horizontal distance in presence of resistance

[songoku]

Homework Statement

A particle of mass m kg is projected under gravity with horizontal and vertical components of velocity U and V. There is horizontal resistance of magnitude mku, where k is constant and u is horizontal component of the velocity at t seconds. No vertical resistance present.
(i) Show horizontal range from point of projection is R = \frac{U}{k} (1 - e^{-kT}) where T = 2V/g
(ii) Find the horizontal distance D traveled before reaching the highest point of its path
(iii) Show that D > R/2

Homework Equations

Integration
Newton's law
Kinematics

The Attempt at a Solution

(i) I did this and got the answer

(ii) For this question, is it only changing the time from T to T/2? So the answer is R = \frac{U}{k} (1 - e^{-\frac{kT}{2}}) ?

(iii) I do not know how to do this one.

\frac{D}{R} = \frac{1 - e^{-\frac{kT}{2}}}{1 - e^{-kT}} and I need to show this will be bigger than 1/2

Thanks

 

[BvU]

ii No
iii Neither

 

[BvU]

Correction no vertical resistance, so: parabola y(t), max at halfway

 

[BvU]

iii can you divide $1-e^ {kt}$ by $1-e^{kt\over 2}$ ?

 

[songoku]

iii can you divide $1-e^ {kt}$ by $1-e^{kt\over 2}$ ?

\frac{D}{R} = \frac{1 - e^{-\frac{kT}{2}}}{1 - e^{-kT}}

= \frac{1 - e^{-\frac{kT}{2}}}{(1 - e^{\frac{-kT}{2}}) (1+ e^{\frac{-kT}{2}})}

= \frac{1}{1+ e^{\frac{-kT}{2}}}

Then how to proceed?

Thanks

 

[ehild]

\frac{D}{R} = \frac{1 - e^{-\frac{kT}{2}}}{1 - e^{-kT}}

= \frac{1 - e^{-\frac{kT}{2}}}{(1 - e^{\frac{-kT}{2}}) (1+ e^{\frac{-kT}{2}})}

= \frac{1}{1+ e^{\frac{-kT}{2}}}

Then how to proceed?

Thanks

can be = e^{\frac{-kT}{2}} greater than 1?

 

[songoku]

can be = e^{\frac{-kT}{2}} greater than 1?

In this question, k can not be negative?

Thanks

 

[BvU]

resistance

No, resistance is always opposed to the motion, so $k$ is positive.

 

[songoku]

Thank you very much for the help BvU and ehild