Finding the Time to Reach Maximum Height with Air Resistance

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In summary, the conversation is about finding the time it takes for a canonball launched with velocity v_0 to reach its maximum height, taking into account air resistance. The correct solution is t=-ln[(v_0/g)k+1]/k, without the minus sign. There was some discussion about the integration and the use of negative signs in the equations, but it was eventually clarified that the minus sign should not be included.
  • #1
captain
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Homework Statement


A canonball is launched up in the air with velocity v_0. There is air resistance, which is equal to kmv (where k is some proportionality constant and m is the mass of the canonball) and is proportional to the canonball's velocity. How long does it take to reach its maximum height.


Homework Equations





The Attempt at a Solution



I got t=ln(g)/k as the time. I would just like to verify if that is the correct solution. g=earth's gravitational acceleration constant.
 
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  • #2
Your answer is incorrect. Perhaps if you showed your working we could point you in the right direction...
 
  • #3
Hootenanny said:
Your answer is incorrect. Perhaps if you showed your working we could point you in the right direction...

i realize that my units are off by a lot. i realized that i get the total acceleration to be
-kv+g=(dv)/(dt), but when i try and integrate it by (dv)/(-kv+g)=dt i get incorrect units. is the g supposed to stay on the other side of the equation. If so how can I integrate it? and after that I am not quite sure about how to use v_0 given to me?
 
  • #4
i would like to verify if this is correct. I checked to make sure if it is in the right units.

t=-ln[(v_0/g)k +1]/k
 
  • #5
captain said:
i would like to verify if this is correct. I checked to make sure if it is in the right units.

t=-ln[(v_0/g)k +1]/k

I'm getting exactly the same thing but without the minus sign.

I think the minus sign shouldn't be there because ln(positivenumber + 1) > 0... so with that minus sign there you'll get a negative time.
 
  • #6
i got it in terms of my integration bounds for the integral with dv. i set the bounds from v to 0
 
  • #7
-kv+g=(dv)/(dt),

should you be using -kv - g... or are you using g = -9.8 instead of g = 9.8?
 
  • #8
i see i was doing it wrong the whole time
 
  • #9
i accidentally made the force in my free body diagram like this:

m(-a)=-f_drag-mg

i thought that since the acceleration was negative with respect to my reference frame, the -a was necessary when the right hand side of the equation took care of that.
thanks for your help.
 

1. What is a 1-d kinematic problem?

A 1-d kinematic problem involves studying the motion of an object in one dimension, typically along a straight line, and using equations to analyze its position, velocity, and acceleration over time.

2. How do you solve a 1-d kinematic problem?

To solve a 1-d kinematic problem, you need to identify the given information, such as initial and final positions, velocities, and accelerations. Then, use the appropriate kinematic equations to find the unknown quantity.

3. What are the four kinematic equations used in 1-d problems?

The four kinematic equations used in 1-d problems are:
1. x = x0 + v0t + 0.5at2
2. v = v0 + at
3. v2 = v02 + 2a(x-x0)
4. x = x0 + 0.5(v + v0)t
where x is the final position, x0 is the initial position, v is the final velocity, v0 is the initial velocity, a is the acceleration, and t is the time.

4. What are the units of measurement for position, velocity, and acceleration in 1-d kinematic problems?

The units of measurement for position in 1-d kinematic problems are meters (m), for velocity are meters per second (m/s), and for acceleration are meters per second squared (m/s2).

5. How is displacement different from distance in 1-d kinematic problems?

In 1-d kinematic problems, displacement refers to the change in position of an object from its initial position to its final position, taking into account both direction and magnitude. Distance, on the other hand, refers to the total length of the path traveled by the object, regardless of direction. Displacement can be positive or negative, while distance is always positive.

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