Projectile motion with air resistance

AI Thread Summary
The discussion focuses on solving a projectile motion problem involving air resistance, where a ball is thrown from a height of 30 meters with an initial velocity of 20 m/s. The equations of motion are set up, but confusion arises during integration and application of initial conditions, leading to negative time values. Participants emphasize the importance of correctly applying initial conditions and adjusting equations for the direction of the drag force when the ball descends. There is a consensus that the integration approach needs refinement to yield accurate results for both maximum height and time of impact. Properly addressing these issues is crucial for solving the problem effectively.
Warr
Messages
119
Reaction score
0
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.
 
Physics news on Phys.org
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
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Back
Top