A stone falls from rest on a cliff A 2nd stone is thrown from the same height 2.4 s later with an initial speed of 47.04 ms They land at the same time How long does it take the 1st stone to land?

Answer 1

Position of an object accelerating(or decelerating) uniformly can be determined using following equation:

x = v_0 + 0.5 * a * t^2, where

x - position,

v_0 - initial speed,

a - acceleration,

t - time since start.

Here, we've got two equations:

h = gt^2 for the first stone

and

h = 47.04 + g(t-2.4)^2 for the second stone.

H and t are the same in both cases, because they fall from the same height and reach the ground at the same time.

When we put the equations together:

gt^2 = 47.04 + g(t-2.4)^2,which simplifies to:

gt^2 = 47.04 + gt^2 - 4.8t + 5.76, and then

4.8t = 52.8, which gives:

t = 11s.

Answer 2

You did such a beautiful job of stating the conditions that apparently you forgot

to ask the question.

We suspect the question is: How high is the cliff above the ground below ?

In return for our being responsible for figuring out what the question is, we're not

going to show all of our work here, or any of the gyrations we went through to get

an answer. We'll just tell you the answer we got.

Of course, if we guessed wrong at the question, then the answer isn't worth much.

We found that the cliff looms 41.468 meters(136.05 feet) above the canyon floor.