Free fall distance?

Question

Ok, I've been working on this and I am stumped. I don't necessarily want the answer, but a point in the right direction:

A free falling object takes 1.50 s to travel the last 29.0 m before it hits the ground. From what height above the ground did it fall [no wind resistance]?

Answer 1

It fell from a height of 36.31... m.

Call the time of fall t, the depth of the ground from the starting point d2, and and the depth of the place where the object was 1.50 s earlier, d1.

In free fall, d2 = 1/2 g t^2, while d1 = 1/2 g (t - 1.50)^2.

(That's because the distance things fall in any time T is 1/2 g T^2. The object has travelled a distance d2 after t s, but only the distance d1 after (t - 1.50) s.)

Then d2 - d1 = 1/2 g (t^2 - (t - 1.50)^2).

That is promising: the SQUARED term in t vanishes! The result is:

29.0 m = 1/2 g (3 t - 2.25) m, so that 3/2 g t = 29.0 + 1/2 g 2.25.

Thus t = 2*29.0 / (3 g) + 2.25 / 3 = 1.97078... + 0.75 = 2.72078... s.

Then the height it fell from was 1/2 g t^2 = 36.31... m.

Live long and prosper.

Answer 2

s = ut + 1/2 gt^2; where s = 29 m, t = 1.5 sec, and g = 9.81 m/sec^2 on Earth's surface.

Solve for u, which is the velocity of the object when it starts its last 29 m of fall.

Then H = s + S; where H is the height of the drop above ground and S is the distance it already fell to reach the u velocity.

From u = gT, we get T, the time the object fell the distance S. Then S = 1/2 gT^2 = 1/2 g(u/g)^2 = 1/2 u^2/g can be found and folded into H = s + S, which is your answer.

Answer 3

If you think about it, just figure out the factors we know for each time period. T1 will be from the moment the object is let go, until T2, and T2 will be the moment from 29m above the ground until it hits the ground.

X1i = ?
X1f = 29 m
V1i = 0 m/s (letting go)
V1f = ?
A1 = 9.8 m/s²
T1 = ?

X2i = X1f = 29 m
X2f = 0 m (the ground)
V2i = V1f = ?
V2f = ?
A2 = 9.8 m/s²
T2 = 1.5 s

If you can figure out what V2i is, then you have your V1f, and you can have 4 variables known to solve the problem.

Answer 4

gravitaional pull is 9.81m/s squared, ie. if you drop something from a height, in the first second it will drop 9.81metres, then the next second it will drop even faster, ie 9.81ms (already travelling at) + an additional 9.81 m/s for the 2nd second, and so on.

you need to work out its speed for the last 1.5s. ie if it travels 29m in 1.5secs then its speed at that point is 20m/s +/-.

you then just need to work it backwards until you reach a point where its velocity is 0. id hazard a guess and say about 50metres.? dont take my answer as correct it was a rough guess. but the above way is the method to work it out. ie work it backwards from its final speed