Projectile motion with air resistance

In summary, the problem being discussed involves a ball being thrown upwards from the top of a building with an initial velocity of 20 m/s in the y direction. The air resistance is given as (1/30)*v and the goal is to find the maximum height and time when the ball hits the ground. The conversation includes different attempts at solving the problem, but there is still confusion and incorrect equations.
  • #1
Warr
120
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.
 
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  • #2
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.
 
  • #3
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:
  • #4
You didn't apply your initial condition properly when you integrated.
 
  • #5
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
 

1. What is projectile motion with air resistance?

Projectile motion with air resistance is a type of motion in which an object is thrown or launched into the air and is affected by the force of air resistance. This force acts opposite to the direction of the object's motion and can cause changes in its trajectory.

2. How does air resistance affect projectile motion?

Air resistance can affect projectile motion by slowing down the object's velocity and causing it to deviate from its original path. This is because the force of air resistance increases as the object moves faster, creating a drag force that acts against the object's motion.

3. How is air resistance calculated in projectile motion?

The force of air resistance on an object in projectile motion is calculated using the equation F = ½ * ρ * v² * A * C, where ρ is the density of air, v is the velocity of the object, A is the cross-sectional area of the object, and C is the drag coefficient which depends on the shape of the object.

4. Can air resistance be ignored in projectile motion?

In most cases, air resistance cannot be ignored in projectile motion as it can significantly affect the trajectory and final position of the object. However, in some situations where the object's velocity is low and the duration of its flight is short, the effects of air resistance may be negligible and can be ignored.

5. How does air resistance change with different objects?

The amount of air resistance experienced by an object in projectile motion depends on its shape, size, and surface area. Objects with larger surface areas, such as a parachute, experience more air resistance compared to smaller and more streamlined objects, like a bullet.

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