Physics problem help please~~~~~~~~~~~?
On a certain day, the speed of sound is 343m/s. If a stone is dropped from a cliff 125m high then how long after the stone is dropped(from rest) will the person on top of the cliff hear the stone strike the ground? (g=9.80m/s2)
I have no idea of how to solve this problem. Can someone explain to me how to do it please?
Comments
Time to drop 125m = sqrt. (2h/9.8) secs., = 5.05 secs.
Time for sound to reach ears = (125/343) = .364 secs.
(5.05 + .364) = 5.414 secs.
Okay, you can determine the time by using (delta)x=1/2a(t)^2
so 125=9.81/2 (t)^2
250= 9.81(t)^2 so divide both sides by G
which is 25.5=t^2
find the square root, then you have 5.1=t
Now that you have The amount of time it actually takes the rock to hit the bottom of the cliff now you have to find how long it takes for the sound of the rock to get back to the person.
you don't have an initial velocity because the sound starts from rest..I think
So you do
x= 1/2(Vf)t
125=1/2 (343) t
So you find t
125x2 =250
so 250= 343t
then divide both sides
0.73=t
now that you have both times you just add it together so 0.73+5.1=5.83 seconds
SO it's 5.83 seconds...I think I may be wrong so just cross reference this result by googling the first sentence and then putting the givens behind it.
Let t = time for the stone to reach the ground
=> 125 = 4.905t^2
=> t = â(125/4.905) = 5.0482 seconds
Time for the sound to reach the top after the stone reached the ground
= 125/343 = 0.3676 sec.
Time of hearing the sound after the sgtone is dropped
= 5.0482 + 0.3676
= 5.4158 sec.
The converse of this problem is challenging.
"Given that the sound is heard 5.4158 seconds after the stone is dropped, what is the height of the cliff."
Try it.
y = v0 * t + 1/2*g*t^2
solve for t with v0 = 0 then t = sqrt(2*y/g)
that's the time for stone to hit ground.
then it's a simple t = y / V to get time for sound to reach back at you.
Add em together. good luck!
find how many seconds it takes to drop. ( hieght/ gravity maybe) then multiply that by speed of sound.