Suppose a rock is dropped into a well 24.5 meters deep.... anyone?

Question

a) Determine the amount of time t it takes for the rock to hit the bottom of the well.
b) Determine the velocity v of the rock as it is hitting the bottom of the well.
c) If the bottom of the well is lined with thick sand so that the rock does not bounce, what is the velocity vf of the rock just after it hits the bottom of the well?

for (a) I got 2.5 seconds..assuming that the free fall speed is 9.8m/s but it seems that this answer is wrong... I can't figure it out.. can anyone help me?

Answer 1

What method are you using to solve this?

that 9.8 is not a velocity, it's an acceleration (or change in velocity). It's units are m/s^2. So you can't use x=vt to solve it since the velocity is actually changing

Well.. let the bottom of the well be our zero position... so the rock starts with an initial height of 24.5.. and under goes an accelleration of -9.8 m/s^2

Using this equation with xi of 24.5 and vi of 0 (since it's dropped)
s(t) = 1/2at^2+vit + xi

So the equation for the rocks motion is
s(t) = -4.9t^2 + 24.5

We want to know the time when s(t) is 0 (rock hits bottom)
-4.9t^2 + 24.5 = 0
-4.9t^2 = -24.5
t^2 = 5
t = sqrt(5) seconds.

Which is a little less than 2.25 seconds..

For the second one we use
vf = at+vi

With a vi of 0 (rock is dropped)... and the time that it hits the ground at.
vf = -9.8 sqrt(5)
roughly -21.9 m/s

c) 0, question implies the rock comes to a stop.. or vf = 0

Answer 2

Hi,
I suspect you did the first part correctly. You’re off a little in the second decimal. So, let’s do it just to make sure you’re on the right track.
Part a:
s = vot + (1/2)at²
s = 0 + (1/2)9.8t² (The initial velocity is zero.)
24.5 = 4.9t²
√(24.5/4.9) = t
2.236 = t

Part b:
Now, the velocity is this:
v= vo + at (In vo the "o" should be a subscript.)
= 0 + 9.8(2.236)
= 21.91 ft/s

Part c:
Frankly, I’m not sure what sort of response they’re trying to elicit with this question. In thick sand the velocity would be very quickly reduced to zero, but of course it cannot immediately be reduced to zero. In fact, the velocity would depend on an exact definition of “just after,” the viscosity of the sand, etc. But since they’ve mentioned bouncing, I suspect that they’re thinking of a velocity of zero.

Hope this helps some.
FE

Answer 3

First we could be attentive to how long it takes for the rock to hit the backside. shall we assume air resistance for the falling rock would be omitted (loose fall) and its preliminary velocity grew to become into 0 whilst dropped. The rock will advance up at g = 9.8 m/s^2 In time dt it falls distance dx : dx = a million/2 g t^2 so 408 = a million/2 9.8 t^2 remedy for t and you come across t ~ 9.a million s The sound wave takes 408 / 318 = a million.28 s to realize the floor. entire time ~ 9.a million + a million.28 = 10.38 s whilst the rock hits the backside its velocity would be v = a t = 9.8 * 9 = 88 m/s, positioned on a helmet!!