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**Question:**"A man steps from the top of a tall building. He falls freely from rest to the ground, a distance of

*h*. He falls a distance of

*h/4*in the last 0.800 s of his fall. Neglecting air resistance, calculate the height

*h*of the building.

I've attempted the solution in a number of different ways, each of which give different answers.

**Solution:**

I have labelled the initial time (i.e. at the instant before he begins to fall) as ##t_{0}##, the time at which he has traveled ##\frac{3}{4}## of the total distance

*h*as ##t_{1}## and the time at which he has traveled the full height

*h*as ##t_{2}##.

Using the equation ##x = x_{0} + v_{0}t + \frac{1}{2}a_{x}t^{2}##, where ##x = h##, ##t = t_{2}## and ##a_{x} = g##.

##x_{0} = 0## as no distance has been traveled whilst the man is still at the top of the building

##v_{0}t = 0## as the initial velocity at the top of the building is zero

Therefore -

## h = h_{0} + v_{0}t_{2} + \frac{1}{2}gt_{2}^{2}##

## h = 0 + 0 + \frac{1}{2}gt_{2}^{2}##

## h = \frac{1}{2}gt_{2}^{2}##

Re-arranging to give -

##t_{2} = \sqrt{\frac{2h}{g}}##

##(t_{2} - t_{1})## = the time to travel ##\frac{1}{4}h = 0.800 s##

##(t_{1})## = the time to travel ##\frac{3}{4}h = \sqrt{\frac{2h}{g}} - 0.800 s##

Substituting the value of ##t_{2}## for ##h## in the latter of these equations gives -

##\frac{3}{4}\sqrt{\frac{2h}{g}} = \sqrt{\frac{2h}{g}} - 0.800 s##

Therefore 0.800 must be equivalent to ##\frac{1}{4}\sqrt{\frac{2h}{g}}##

##0.800 = \frac{1}{4}\sqrt{\frac{2h}{g}}##

##3.20 = \sqrt{\frac{2h}{g}}##

##h = \frac{3.20^{2}g}{2}##

##h = 50.2 m##

I believe that my mistake is in assuming that ##(t_{1}) = \sqrt{\frac{2h}{g}} - 0.800 s## is equivalent to ##\frac{3}{4}h\sqrt{\frac{2h}{g}}##, as the man's velocity increases with constant acceleration and therefore ##\frac{3}{4}## of ##t_{2}## does not correspond to having traveled ##\frac{3}{4}h##.

However, I'm uncertain how else to approach the problem. I have tried other methods involving heavier use of the equations of motion with constant acceleration and have twice calculated

*h*as ≈ 12 m, but this would not be a tall building! (I realise that the actual answer does not always have to accurately reflect the real-world scenario described in the question, but I believe I should use it as guidance in evaluating my answer.)

Any help and/or advice would be greatly appreciated.

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