# Projectile with linear drag

• BearY
In summary, the beach ball is thrown upwards with initial speed v0 and the drag force is given by Fd = −mαv for a drag coefficient α and speed v. The height h reached can be expressed in terms of the final speed when the ball strikes the ground vf. Finally, the total time taken for the ball to travel up and back down is given by T = (v0 + vf )/g.f

## Homework Statement

A beach ball is thrown upwards with initial speed v0. The drag force is given by Fd = −mαv for a drag coefficient α and speed v.
(a) the height h reached
(b) Show that the height may be expressed in terms of the final speed when the ball strikes the ground vf
(c)Finally show that the total time taken for the ball to travel up and back down is given by T = (v0 + vf )/g.

## The Attempt at a Solution

I solved DE for V
$$\dot{V} = \alpha V}+{g}$$
get $$t=\frac{1}{\alpha}ln(\frac{\alpha V_0 +g}{g})$$
and DE for h $$\ddot h -\alpha \dot h = g$$
get $$h = \frac{gln(\frac{\alpha V_0 +g}{g})}{\alpha ^2}+C_1\frac{g}{\alpha V_0 +g}+C_2$$
There is the problem, I can only find initiail value of ##t=0##, which does not solve the 2 constants. Is there another point of time I should use? Or there is something I did wrong?

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When an object is thrown straight up, the acceleration due to drag and the acceleration of gravity are in the same direction. This means that they must have the same sign in your differential equation.

• BearY
When an object is thrown straight up, the acceleration due to drag and the acceleration of gravity are in the same direction. This means that they must have the same sign in your differential equation.
I am sorry I forgot to put vector signs.

This is a one-dimensional vector equation. Regardless of whether the positive axis is "up" or "down" the two accelerations are in the same direction, so they must have the same sign, both positive or both negative. Also, note that symbol V stands for speed as indicated in the problem; it's not a velocity. I suggest you think of it as speed otherwise you might be led astray with too many or too few negative signs. You will need a slightly different differential equation for the return trip, of course.

• BearY

## Homework Statement

$$\dot{\vec{V}} = -\alpha \vec{V} +{g}$$
get $$t=\frac{1}{\alpha}ln(\frac{-\alpha \vec{V_0} +g}{g})$$
and DE for h $$\ddot h +\alpha \dot h = g$$
get $$h = \frac{gln(\frac{-\alpha V_0 +g}{g})}{\alpha ^2}+C_1\frac{g}{-\alpha V_0 +g}+C_2$$
There is the problem, I can only find initiail value of ##t=0##, which does not solve the 2 constants. Is there another point of time I should use? Or there is something I did wrong?
Adding arrows over symbols doesn't make it right. See my comments in #4.

• BearY
Adding arrows over symbols doesn't make it right. See my comments in #4.
I will use V as speed and change the DE. But if V is pointing up, wouldn't ##\alpha V## be pointing up as well?

I will use V as speed and change the DE. But if V is pointing up, wouldn't ##\alpha V## be pointing up as well?
Yes, ##\alpha v## points up, but that's not the drag acceleration when the object is moving up. When the object is moving up, the drag acceleration is ##a_d=-\alpha v## and points down. Also, it looks like you did not integrate the diff eq. in v correctly. You need to get v as a function of time first by doing a definite integral noting that at t = 0 the speed is v = v0 (lower limits) and at t = t the speed is v = v (upper limits).

• BearY
Yes, ##\alpha v## points up, but that's not the drag acceleration when the object is moving up. When the object is moving up, the drag acceleration is ##a_d=-\alpha v## and points down. Also, it looks like you did not integrate the diff eq. in v correctly. You need to get v as a function of time first by doing a definite integral noting that at t = 0 the speed is v = v0 (lower limits) and at t = t the speed is v = v (upper limits).
That's what I thought at first, drag is always opposite to V, so I thought putting a minus sign there makes it equation of motion of the entire motion. But now I am using V as speed. This is my new solution to the DE
$$\frac{dv}{dt}=-\alpha V -g$$$$\frac{1}{-\alpha} ln(-\alpha V -g)=t+C$$and use t=0 v=v_0, we have
$$V=\frac{e^{\alpha t + \alpha C}+g}{\alpha}$$
$$C = \frac{1}{\alpha}ln(\alpha V_0 +g)$$
Miss clicked was going to preview
and use t=t_1 v=0, we have
$$t_1=\frac{1}{\alpha}ln(\frac{g}{\alpha V_0 +g})$$

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That's what I thought at first, drag is always opposite to V, so I thought putting a minus sign there makes it equation of motion of the entire motion. But now I am using V as speed. This is my new solution to the DE
$$\frac{dv}{dt}=\alpha V +g$$
Look at the right hand side of the equation. You have two terms, both of which are positive. This means that the rate of change of the speed is positive which means that the object is moving faster as it goes farther up. Does this make sense? You want both quantities on the right hand side negative because both the acceleration of gravity and drag work together to reduce the speed.

Also, the presence of integration constant ##C## shows that you did not use definite integrals. Use limits when you integrate as I indicated in post #7 and the integration constants will take care of themselves.

• BearY
Look at the right hand side of the equation. You have two terms, both of which are positive. This means that the rate of change of the speed is positive which means that the object is moving faster as it goes farther up. Does this make sense? You want both quantities on the right hand side negative because both the acceleration of gravity and drag work together to reduce the speed.

Also, the presence of integration constant ##C## shows that you did not use definite integrals. Use limits when you integrate as I indicated in post #7 and the integration constants will take care of themselves.
Should I integrate right-hand side from ##V_0## to ##0##? I think I am wrong because that gives the same results.

Look at the right hand side of the equation. You have two terms, both of which are positive. This means that the rate of change of the speed is positive which means that the object is moving faster as it goes farther up. Does this make sense? You want both quantities on the right hand side negative because both the acceleration of gravity and drag work together to reduce the speed.

Also, the presence of integration constant ##C## shows that you did not use definite integrals. Use limits when you integrate as I indicated in post #7 and the integration constants will take care of themselves.
Wait it's not exactly the same the result is $$\frac{1}{\alpha}ln(\frac{\alpha v_0+g}{g})$$

Should I integrate right-hand side from ##V_0## to ##0##? I think I am wrong because that gives the same results.
Look at this example when ##\alpha =0## that gives a familiar equation.
$$\frac{dv}{dt}=-g~~~\rightarrow ~dv=-gdt$$ $$\int_{v_0}^{v}dv=-g\int_0^t dt~~~\rightarrow ~v-v_0=-gt~~~\rightarrow ~v=v_0-gt$$Note how the upper and lower limits are lined up; ##v_0## is the speed at ##t## and ##v## is the speed at arbitrary time ##t##. In other words, it is the speed at any time ##t##, or ##v(t)##.

Sorry, I have to leave now because it's late where I am.

• BearY
Wait it's not exactly the same the result is $$\frac{1}{\alpha}ln(\frac{\alpha v_0+g}{g})$$
Yes, that is the correct time ##t_{up}## for the up trip. However, you need to find ##v(t)## as I indicated before because you need to do a second definite integral of the form $$h(t_{up})=\int_0^{t_{up}}v(t)~dt.$$