Question

At t = 0 , a stone is dropped from a cliff above a lake; 1.5 seconds later another stone is thrown downward from the same point with an initial speed of 32 m/s . Both stones hit the water at the same instant. Find the height of the Cliff

Answer 1

Does the second stone have an acceleration of -g also?

Answer 2

For the first equation I get \[X _{f} = X _{i}\] is this correct?

Answer 3

@Shane_B Can you help please?

Answer 4

@CliffSedge ?

Answer 5

Did you draw a diagram first?

Answer 6

Yeah , I'm trying to see what would happen when the first stone is at t = 1.6. Not sure if that'll help though

Answer 7

No, consider that both are going the same distance, but one is doing it in 1.5s less time.

Answer 8

You can try setting up two x=vt+0.5t^2 equations where x is the same for both and solve for t.

Answer 9

oops, x=vt+0.5(a)t^2 [forgot the acceleration]

Answer 10

So is x = ( x final - x initial )?

Answer 11

Yes, but x final = 0. (your choice)

Answer 12

Maybe this will help:\[\frac{1}{2} gt^2=(32m/s)(t-1.5)+\frac{1}{2} gt^2\]\[t=\frac{3}{2}s\]

Answer 13

Hmm, remember that the second one happened in 1.5s less time.

Answer 14

What about the second term?

Answer 15

Yea...brainfart

Corrected:\[\frac{1}{2} gt^2=(32m/s)(t-1.5s)+\frac{1}{2} g(t-1.5s)^2\]\[t=2.13728s\]

Answer 16

But at 1.5 seconds before t = 0 , But why does the equation start with (1/2)gt^2? Wouldn't it equal zero?

Answer 17

The left side is the stone that is dropped...the right side is the stone that is thrown downward at 32m/s, 1.5s later.

Answer 18

If they hit at the same time, the distances must be equal. All you need to do now is plug t into either side of the equation to get the distance.

Answer 19

Ok ok . Sorry but idk if I'm not seeeing it right but the stone dropped was dropped at t = 0 though

Answer 20

Yes, that's your base line so nothing changes in the basic equation.

Answer 21

Yes...let's look at it one by one. The stone that was dropped at t=0s. The distance it had to fall would be:\[d=\frac{1}{2}gt^2\]The second stone was thrown 1.5s later...and fell the same distance in basically 1.5s less time. Since it had an initial velocity the distance for that stone would be: \[d=V_0(t-1.5s)+\frac{1}{2}g(t-1.5s)^2\]If they hit the water at the same instant, these distances must be equal so you can set them equal to solve for t...which is what i did above.

Answer 22

Once you know t you can solve for the distance:\[d=\frac{1}{2}gt^2=\frac{1}{2}(9.8m/s^2)(2.13728s)^2=22.38m\]

Answer 23

Ok ok I can see it. So the X initial is basically a baseline also?

Answer 24

I didn't use any formulas with Xi or Xf...just d=V0t+(1/2)gt^2

Answer 25

I guess you could say d=Xf-Xi...but there's no point :)

Answer 26

Oh ok I see what you mean! Thank you @Shane_B You've really been a lot of help.

Answer 27

Np, good luck :)