# Air resistance (pebble throwing)

1. Mar 26, 2007

### jojoking

i have solved the question of a pebble thrown upwards without air resistance
speed 10m/s mass 0.02kg ma=w, w=-mg

man on cliff 50 feet high throws pebble upwards so it finishes in sea,
what height does it reach v=0 t = 10/g approx =1
since motion started at t =0 & duration of time 1s
x = 100/g - 100/2g is approx 5.1m so maximum height 55.1 m

BUT
i now need to find height with air resistance 0.008 v
any pointers to x value warmly recieved
cheersjk

2. Mar 26, 2007

### HallsofIvy

Staff Emeritus
Again, mass times acceleration equals force.

Without air resistance, the only force its weight: -mg so you have
$$m\frac{dv}{dt}= -mg$$
The easy solution to that is v= -gt+ C. Since the initial velocity is 10 m/s
v(t)= dx/dt= -gt+ 10. Integrate again to get x as a function of t. Solve v(t)= -gt+ 10 to find when it stops going up, put that value of t into the x fuction to find the highest point.

I said that to review what you must have done. Now, with air resistance, the force is -mg- 0.008v (negative because friction force always opposes v).

$$m\frac{dv}{dt}= -mg- 0.008v$$
You can rewrite that as
[tex]\frac{mdv}{mg+0.008v}= -dt[/itex]
and integrate both sides to find v as a function of t. From there the rest of the problem is the same.

3. Mar 26, 2007

### jojoking

i understand your reasoning, thank you

cheers jk