Projectile motion with air resistance

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Homework Help Overview

The problem involves projectile motion with air resistance, specifically focusing on a ball thrown vertically from a height of 30 meters. The original poster presents equations of motion that include gravitational force and a drag force proportional to velocity.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore different methods for integrating the equations of motion, questioning the application of initial conditions and the behavior of the drag force during ascent and descent.

Discussion Status

There is ongoing exploration of the equations and integration techniques, with some participants suggesting that initial conditions may not have been applied correctly. Multiple interpretations of the drag force's effect on motion are being discussed.

Contextual Notes

Participants note confusion regarding the setup of the equations and the implications of air resistance on the trajectory, particularly when calculating the time to reach maximum height and the subsequent descent.

Warr
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Ok, in this problem, only the y direction is dealt with

Here is the problem

You are standing on top of a building, 30 m above the ground. You throw a ball (m = 0.15 kg) with an initial velocity or 20 m/s (in the y direction). Air resistance is given as (1/30)*v (yet again only considering the y direction because range is not an issue here).

a) find the maximum height of the ball
b) find the time t when the ball hits the ground 30 m below you

Ok so I set it up like this

mv' = -mg - v(1/30)
v' = -g - v(1/30m)

from here I get kind of confused.

I tried this

v' + v(1/30m) = -g
(e^(t/30m)v)' = -ge^(t/30m)
integrating:
e^(t/30m)v = -30mge^(t/30m) + c
v = -30mg + c/(e^(t/30m))

I'm pretty sure I'm wrong by here. Help would be appreciated.
 
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That will give you the time it takes to reach the top of the trajectory. Don't forget that the direction of the drag force reverses on the way down so you'll have a different equation to solve for that part.
 
well..I don't even think I did it right

When I try to solve for t for the intial codition of V(0) = 20, and then set V = 0 in order to solve for the time at which the ball has reached its peak, I get a negative time. So I assume my equation is wrong..dunno what to do though
 
Last edited:
You didn't apply your initial condition properly when you integrated.
 
I tried everything I could, including

v' = -g - kv where k = (1/30m)
v'/(-g-kv) = 1
v'/(g/k + v) = -k
integrating with following limits
ln(g/k + v)|from v_o to v = -kt (from 0 to t)
ln(g/k + v) - ln(g/k + v_o) = -kt
ln [(g/k + v)/(g/k+v_o)] = -kt
g/k + v = e^(-kt)(g/k + v_o)
v = e^(-kt)(g/k+v_o) - g/k
Is it right
If it is, then integrating again would this also be right
y-y_o = -(1/k)e^(-kt)(g/k+v_o) - (g/k)t
y = -(1/k)e^(-kt)(g/k+v_o) - (g/k)t + y_o

I think this 'looks' right but when I try to calculate the displacement for the time I calculated where the ball should have a velocity of 0 (about 1.6835 s), by plugging into the second equation, I get a very negative value..Thanks in advance
 

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