Can you help me with a physics kinematics problem, it's NOT easy?

Question

A stone is dropped from a tall building. At some point in the stone's fall, it passes a window that is 2.2m tall. It takes the stone 0.28s to go past the window from top to bottom. How far up past the window was the stone initially dropped from?

Answer 1

Let the stone take a time ' t ' seconds to fall to the BOTTOM of the window. Then it will have been dropped for a time
(t - 0.28)s when it passes the TOP of the window.

Since the stone falls a distance 1/2 g T^2 in time T, we can express the height of the window as the DIFFERENCE between how far it falls in times t and (t - 0.28) seconds:

Thus 1/2 g [t^2 - (t - 0.28)^2] = 2.20m, that is

0.56 t - 0.0784 = 4.40/g = 0.44898...

Thus t = 0.52738.../0.56 = 0.94175... s.

So 1/2 g t^2 = 4.35... m

So the stone was dropped from a height of 4.35... m above the bottom of the window, that is from 2.15 m above the TOP of the window.

Live long and prosper.

P.S. The two answers below are clearly wrong.

1. The first of them, by 'RJJ,' asserts:

"This velocity of 7.86 Meters/sec is for all intents and purposes the final velocity for this scenario."

But it simply ISN'T. It's only the average velocity as it passes the window. Clearly the velocity at the bottom of the window is GREATER.

2. The mistake by 'eyeonthe screen' is a little more subtle. He asserts:

"We can assume the end velodity v = u + gT to equal the average velocity at s = h/2 = 1.1 m, which is the middle of the window height. "

But we CAN'T assume that. The speed goes up linearly with TIME, not with DISTANCE. (In fact the SQUARE of the speed goes up with distance, as one can see from v^2 = u^2 + 2gd, where d is the distance dropped.)

However, since this wrong assumption is closer to the truth than the previous one, the answer is closer to mine.

The bottom line ? : My argument is the only correct one so far.

Answer 2

OK, I'm doing this because you have two different answers above. Here's a third one.

Time to fall past the window = t = .28 sec and window height = h = 2.2 m.

Average velocity past the window = h/t = 2.2/.28 = 7.86 m/sec = v.

We can assume the end velodity v = u + gT to equal the average velocity at s = h/2 = 1.1 m, which is the middle of the window height. Thus, end velocity after falling from the roof T seconds = v = u + gT down to mid-way on the window height and T = v/g, the time to fall from the roof top assuming the initial velocity u = 0.

Therefore the stone dropped H = s + S = 1/2 gT^2 from the roof top; where S = distance from the roof to the top of the window, the answer you are looking for.

S = H - s = 1/2 g(v/g)^2 - s = 1/2 v^2/g - s = 1/2 (7.86)^2/9.81 - 1.1 = 3.15 - 1.1 = 2.05 m above the top window sill. So there you have it, my answer is different from the other two. Take your choice.

Answer 3

Acceleration of gravity = 9.8 meters/sec/sec
Window height = 2.2 meters
Time to pass window = 0.28 sec

The stone will continue to accelerate until it reaches terminal velocity. We will assume that the stone has not yet reached terminal velocity.

First determine the velocity (meters/sec) of the stone as it passes the window:

Velocity = (change in distance)/(change in time)
Velocity = (2.2 Meters)/(0.28 sec)
Velocity = 7.86 Meters/sec

This velocity of 7.86 Meters/sec is for all intents and purposes the final velocity for this scenario. If so:

Final velocity = Initial velocity +(Accel due to gravity)x(time change)
7.86 M/sec = 0+(9.8 M/sec/sec)x(time change)

Solve for time change; time change = 0.8 sec
Which give us the total time the stone fell to the bottom edge of the window.
Hence, the distance the stone has fallen once it has reached the bottom of the window opening is:

displacement = 0.5(9.8 M/sec/sec)(0.8)^2
displacement = 3.136 meters

the stone was (3.136 - 2.2) or 0.936 meters above the window.