a second physics question?

Question

At t = 0, a stone is dropped from a cliff above a lake; 1.8 seconds later another stone is thrown downward from the same point with an initial speed of 41 m/s. Both stones hit the water at the same instant. Find the height of the cliff.
m

Answer 1

Use Excel and create a column with time in increments of 0.1 seconds that is 100 cells deep. In the second columns, calculate v=gt for each time interval. In the third column, calculate d=1/2gt for each time increment. Imagine where each stone is, depending on time, within the spectrum. At some point, the second stone will catch up with the first which is the only time there are side/by side which must be the time they hit the water.

The answer is approximately 10.2 m

Answer 2

there's plenty of information to solve this problem, don't listen to craptasticness. second, using excel to solve this is idiotic and won't give you a precise answer. this is physics, not business class.

the equation you're looking for is:

D = v(0) * T + (0.5) A * T * T

You now have two equations since you know V(0) in both cases (0 for the first, 41 for the second). You also have two variables, D and T1 (because T in the second equation (T2) is actually T1 + 1.8... don't forget that when you come up with the equation for the second rock). You can then make a substitution and solve to get both variables.

Answer 3

No answer, too many variables.

Wind speed, weight of stones, weather conditions, height of swell, mass of stone, terminal velocities, etc etc

Answer 4

Sorry but you didn't mention which universe you are in. Its obviously one where the laws of Physics from this universe do not apply.

Answer 5

You need to give more info man! What answer are you looking for here?

Answer 6

Not enough information.