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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. please explain in detail

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.

please explain in detail

Grade:11

3 Answers

suryakanth AskiitiansExpert-IITB
105 Points
13 years ago

Dear Shubham,

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

Please feel free to ask your queries here. We are all IITians and here to help you in your IIT JEE preparation.

All the best.

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Askiitians Expert

Suryakanth –IITB

vikas askiitian expert
509 Points
13 years ago

let the total time of flight is Tand H be the height of tower.....

 using s=ut+at^2/2

          H=gT^2 /2 ..........1                            (a=g ,u=0)

let total distance covered upto T-1 second be h

  so h=gt^2 /2

       h =g(T-1)^2 /2...........2

  total distance travelled in last 1 second is H-h

    H-h=g(2T-1)/2

   H-h =16H/25   (given)

 so 16H/25=g(2T-1)/2.........3

  after solving  1 and 3

we get a quadratic eq whose roots are   T = 5/8,5/2

 after putting value of T we get H=125/4 ,125/64

 

 

aditya bhardwaj
33 Points
13 years ago

let the height of the tower be h ms then in last sec height=16/25h ms now s= ut 1/2at square time= 1sec then,

16/25 h = U +1/2g =u+5

16/25h-5=u then,u=-109/25h m/s

now for total distance vsquare-usquare=2as

0square-[-109/25h]square=2gh

-1181/25h=20

h=-10or 10m

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