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March 29, 2015

March 29, 2015

Posted by **Jess** on Sunday, September 9, 2007 at 1:12pm.

t takes 6 seconds for a stone to fall to the bottom of a mine shaft. How deep is the shaft? (I know I have to use this formual d=vinitial+1/2gt^2, I plug in these numbers d=0+1/2(9.8m/s^2)(6^2 s= 176.4 m, is this right?)

If a projectile is fired straight up at a speed of 10 m/s, the total time to return to its starting position is about (I know I have to use this formula, hfinal=hinitial+vinitial *time+ 1/2gt^2, these are the numbers I plugged in hfinal=0+0*10m/s+1/2(9.8m/s^2(10m/s)^2=490 s, when I suppose to get 2 seconds for my answer.

Ten seconds after starting from rest, an object falling freely downward will have a speed of about ? (I know I have to use this formual vfinal=gt, here is what I plugged in vfinal=(9.8m/s^2)(10 s)=98 m, when I am suppose to get 1000 m/s

A man leans over the edge of a cliff and throws a rock upward at 4.9m/s. How far below the level from which it was thrown is the rock 2 seconds later? ( would the formula I use be d=vt?

- Physics ASAP -
**bobpursley**, Sunday, September 9, 2007 at 1:43pmyes on the stone falling.

No on the second.

Hfinal=hinitial+ vinitial*t -4.9t^2

0=0+10t-4.9t^2 solve for t

The third: vfinal=gt= 9.8m/s^2*10sec=98m/s. It wont be 1000m/s

The final:

Hfinal=Hinitial+Vi*t - 4.9t2

Let Hintialal be 0

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