Math Is Fun Forum
  Discussion about math, puzzles, games and fun.   Useful symbols: √ ∞ ≠ ≤ ≥ ≈ ⇒ ∈ Δ θ ∴ ∑ ∫ π -




Not registered yet?

  • Index
  •  » Help Me !
  •  » Antiderivative problem...should be pretty easy

Post a reply

Go back

Write your message and submit
:) :| :( :D :o ;) :/ :P :lol: :mad: :rolleyes: :cool: | :dizzy :eek :kiss :roflol :rolleyes :shame :down :up :touched :sleep :wave :swear :tongue :what :faint :dunno

Go back

Topic review (newest first)

2005-11-26 05:34:19

1 inch stone reaches terminal velocity at about 23.93m/s (m=18.9g)
4 inch stone reaches terminal velocity at about 47.86m/s (m=1.2kg)

Our example would suggest that this stone would be traveling at 79.2m/s
A 10.9 inch rock with a mass of 24.74kg would be necessary to achieve this velocity!
I would not recommend throwing a 55 pound anything from that height.

I thought that I would drop a line to show how dramatic air resistance is in everyday situations.

My contribution for anyone that cares:

     terminal velocity = {[8gr(density of object)] / 3(density of medium)}^(1/2)

where: g = -9.8m/s
           r =  radius of object
           object must be smooth and round for this formula

     The above stone calculations were made with these densities.

     density of object = 2200kg/m^3 (based on the average density of the earth's crust)
     density of air = 1.275kg/m^3 (sea level)

2005-11-25 07:45:46

Hi irspow. Good posts. Would you like to become a Member?

2005-11-25 04:23:30


     Air resistance would not be small in this case unless the stone was much more dense than normal rock because the v^2 term in the air resistance formula would be very large nearing impact versus the force of gravitation.
That is nearly 12000(cross sectional area of object) newtons of resistance for a perfectly smooth round surface with zero humidity!  I'm thinking this object would have reached terminal velocity well short of impact.
     That would change both answers significantly.

2005-11-25 03:54:02

Ooh, ganesh missed the initial velocity as well. He even said 'u=0'. He must have been having an off day.

Air resistance makes the whole thing more complicated and in most cases it is too negligible for it to be worth considering, but you're right in that it means that you shouldn't worry about accuracy too much because you're being inaccurate anyway.

I had a physics teacher who kept telling us to ignore air resistance and I kept imagining her obituary:

"A teacher was tragically killed yesterday when she went parachuting, and air resistance ignored her."

2005-11-25 02:16:37

Actually, Ganesh forgot about the initial velocity that the stone was thrown.

2005-11-25 02:15:26

623.097 ft @ 5s
impact at 7.927s
Be careful of the constants being used, 32.15 ft/s^2 makes a notable difference.
Although ignoring air resistance makes this whole discussion moot doesn't it?

2005-11-23 04:22:07

I think ganesh forgot about the '/2' bit in his formula, so got the distance travelled in A double what it should have been, and that error carried forward into B.

2005-11-22 15:35:07

Thanks, I got the answer though.
It came out to be :
A. 625 
B.  7.94618

2005-11-22 15:21:53

S = ut + at/2, u=0
S = (32)(5)= 32*25 = 800
Therefore, after 5 seconds, the stone is (1050-800)=250 feet above the ground.
When S=1050, it hits the ground.
Therefore, 1050=(32)*(t)
t = 1050/32 = 32.8125
t = 5.7822 Seconds (approximately)

2005-11-22 14:32:54

A stone is thrown straight down from the edge of a roof, 1050 feet above the ground, at a speed of 5 feet per second.

A. Remembering that the acceleration due to gravity is -32 feet per second squared, how high is the stone 5 seconds later?
B. At what time does the stone hit the ground?

can anyone help me out with this problem??? it should be easy but i've been stuck on it for a while

Board footer

Powered by FluxBB