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A body falls freely from the top of a tower and during the last second of its flight,it falls 16/25th of the whole distance. Find the height of the tower.

A body falls freely from the top of a tower and during the last second of its flight,it falls 16/25th of the whole distance. Find the height of the tower.

Grade:11

1 Answers

Arun
25750 Points
5 years ago
First, the total height the object falls from may be expressed as:

h = 0.5gt1²---------->(1)

Here, t1represents the TOTAL time the object takes to fall through h.

Now the distance it falls through from rest until it has 1.00s to fall is:

h - 16h/25 = 0.5gt2²
9h/25 = 0.5gt2²------------->(2)

Since we know that that it takes 1.00s to fall part of h, we can say that:

t1 - t2 = 1.00s
t2 = t1 - 1.00s------>(3)

We can eliminate t2by subbing (1) and (3) into (2):

9h/25 = 0.5gt2²
9(0.5gt1²)/25 = 0.5g(t1 - 1.00s)²
0.18gt1² = 0.5gt1² - 1.00gt1+ 0.5g
0.32gt1² - 1.00gt1+ 0.5g = 0

Plug in 9.80 for g (ignore units for now) to get:

3.136t1² - 9.80t1+ 4.90 = 0

From the quadratic formula:

t1 = -b ± √[b² - 4ac] / 2a
= -(-9.80) ± √[(-9.80)² - 4(3.136)(4.90)] / 2(3.136)
t1= 2.5 or t1= 0.625

Because both roots are positive it is not immediately apparent which one makes sense. We can plug them both individually into (3) and if one value for t2is negative, choose the other:


t2 = t1 - 1.00s
= 2.5 - 1.00s
= 1.5s

It's obvious that the other root (0.625s) leads to a negative number, so the time t1 must be 2.5s, the total time for the object to fall.

The distance fallen is easy, as you only need put the total time t1 into (1):

h = 0.5gt1²
= 0.5(9.80m/s²)(2.5s)²
= 31m

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