Vertical flight with air resistance

[brotherbobby]

Homework Statement

A ball is thrown vertically up with some initial velocity. If the resistance due to air (air drag) creates an additional acceleration equal to a tenth of the acceleration due to gravity ($g = 10\; \text{m/s}^2$) :
(a) calculate the percentage change in the ball's total time of flight,
(b) argue whether the ball takes more or less time to be in flight due to air resistance.

Relevant Equations

Total time of flight for a ball thrown vertically up with a velocity $v_0$ is $T = \frac{2v_0}{g}$ where $g$ is the acceleration due to gravity.

I suppose the trick in this question is to realize that the drag acts in opposite directions when the ball ascends and descends and that the ball actually takes less time to rise and more time to fall than normally. I make a small sketch of the problem alongside.

Attempt : The total time of flight in free space : $T = \frac{2v_0}{g}$.
On its way up in air, $T'_{\text{up}} = \frac{v_0}{g+g_a} = \frac{v_0}{g+\frac{g}{10}} = \frac{v_0}{\frac{11g}{10}} = \frac{10 v_0}{11 g}$.
On its way down in air, $T'_{\text{down}} = \frac{v_0}{g-g_a} = \frac{v_0}{g-\frac{g}{10}} = \frac{v_0}{\frac{9g}{10}} = \frac{10 v_0}{9 g}$.
Total time of flight in air : $T' = \frac{10v_0}{g} \left(\frac{1}{9} + \frac{1}{11}\right) = \frac{10v_0}{g} \frac{20}{99} = \frac{200}{99} \frac{v_0}{g}$.
Difference in times : $\Delta T = T'-T = \frac{v_0}{g}\left( \frac{200}{99} - 2 \right) = \frac{2}{99} \frac{v_0}{g} = \frac{1}{99}T$.

(a) The percentage change in the ball's time of flight : $\Delta T$% = $\frac{\Delta T}{T}\times 100 = \frac{100}{99} = \boxed{1.01\%}$.

(b) The ball takes more time in flight due to air resistance since $T'>T$. Of course it travels the same distance either way. It takes less time to rise and a greater time to fall than without air resistance - facts which are both counter intuitive. Exactly as to why $T'>T$ qualitatively, I do not know. Clearly the (absolute value) of acceleration on its way up ($\frac{11g}{10}$) exceeds $g$ by a value $\left(\frac{g}{10} \right)$ which is the same as the amount by which $g$ exceeds it on its way down : $g-\frac{9g}{10} =\frac{g}{10}$. By that argument, the two times should make up when they add and the ball should have the same times of flight with or without air, but it does not.

Any help would be welcome. I suppose I have got the body of the argument correct, but not the tricky bit towards the end where I can't explain why is $T'>T$ qualitatively.

 

[kuruman]

The ball takes more time in flight due to air resistance since . Of course it travels the same distance either way.

Part (a) is good.
In (b) what does "either way" mean? Are you saying that, if thrown with the same initial velocity, the ball will rise to the same height regardless of whether there is resistance or not?

 

[Lnewqban]

You could easily visualize what is happening, by plotting height versus time and velocity versus time graphs.

 

[kuruman]

In part (a) you miscalculated the time for the down trip with air resistance. When you wrote $T'_{\text{down}} = \frac{v_0}{g-g_a} = \frac{v_0}{g-\frac{g}{10}} = \frac{v_0}{\frac{9g}{10}} = \frac{10 v_0}{9 g}$, you assumed that the ball that starts from maximum height with speed zero will reach the ground with the same speed as the launching speed $v_0$. Does that sound right?

 

[brotherbobby]

In part (a) you miscalculated the time for the down trip with air resistance. When you wrote $T'_{\text{down}} = \frac{v_0}{g-g_a} = \frac{v_0}{g-\frac{g}{10}} = \frac{v_0}{\frac{9g}{10}} = \frac{10 v_0}{9 g}$, you assumed that the ball that starts from maximum height with speed zero will reach the ground with the same speed as the launching speed $v_0$. Does that sound right?

Yes, the ball starts from maximum height with speed 0. Let's see based on what I have done, what would its final speed be. $v_F = (g-g_a)T'_{\text{down}} =\frac{9g}{10} \times \frac{10v_0}{g} = v_0!$. Sorry, no that doesn't sound right : $v_F < v_0$ by reason. I'd have to rethink my solution. Many thanks.

 

[kuruman]

Yes, the ball starts from maximum height with speed 0. Let's see based on what I have done, what would its final speed be. $v_F = (g-g_a)T'_{\text{down}} =\frac{9g}{10} \times \frac{10v_0}{g} = v_0!$. Sorry, no that doesn't sound right : $v_F < v_0$ by reason. I'd have to rethink my solution. Many thanks.

Don't forget that "what goes up must come down". What is the maximum height with air resistance?

 

[brotherbobby]

Part (a) is good.
In (b) what does "either way" mean? Are you saying that, if thrown with the same initial velocity, the ball will rise to the same height regardless of whether there is resistance or not?

No, "either way" meaning that the ball travels the same distance on its way up as it does on its way down. Of course the distance it travels will depend on whether there is air resistance or not - with air resistance, the ball will reach a lower height.

 

[brotherbobby]

Don't forget that "what goes up must come down". What is the maximum height with air resistance?

With air resistance, the maximum height $H = \frac{v_0^2}{\frac{2\times 11g}{10}} = \frac{5v_0^2}{11g}$.

 

[kuruman]

That is correct. Now you need to find how long it takes to go back down that distance.

 

[Lnewqban]

Equal upwards and downwards distances, different total accelerations each way, result in dissimilar velocities at ground level, as well as distinct flying times.

 

[brotherbobby]

That is correct. Now you need to find how long it takes to go back down that distance.

Sorry for coming in late. I'd like to do the calculations again where all results above will be summarised.

Attempt : (1) Without air resistance, the total time of flight is given by $T = \frac{2v_0}{g}$.

(2) With air resistance, the acceleration on the way up is $a_{\text{U}}= g_a+g = \frac{g}{10}+g = \frac{11g}{10}$ (acting down) while on the way down $a_{\text D} = g- \frac{g}{10} = \frac{9g}{10}$ also acting down.

(a) The time of flight on the way up (with air resistance): $T'_U = \frac{v_0}{11g/10} = \frac{10}{11}\frac{v_0}{g}$.
The height scaled on the way up : $H = \frac{v_0^2}{2\times \frac{11g}{10}} = \frac{5v_0^2}{11g}$ .

(b) Travelling this height down, the time taken by the ball with air resistance satisfies $H = \frac{1}{2}\times \frac{9g}{10} {T'_{d}}^2$ from where we get $T'_d = \frac{10}{\sqrt {99}} \frac{v_0}{g}$ after some algebra.

Hence, the total time of flight $T' = \frac{10}{\sqrt{11}}\left(\frac{1}{\sqrt{11}} + \frac{1}{3} \right) \frac{v_0}{g} \approx \boxed{1.91 \frac{v_0}{g}} < T$. Hence, with air resistance, the ball goes up and comes down sooner, but it also travels a less height.

Please let me know if my attempt is correct. I have ommited some algebra in between.

 

[kuruman]

So far so good. Everything agrees with what I have.