# A body is dropped from the top of a tower through 40 m during the last two seconds of its fall.Waht is the height of the tower. g=10 ms-1

u=0

h= 40 m

g= 10m/s2

time = 2 sec

h=ut+1/2gt2

h= 0xt + 1/2x10x2

h= 1/2 x10x2

h= 1/2 x20

• -46

No ans is 45.

I have a question to you If a bodt travels 40m during the last 2 seconds of the jouney  HOW CAN HEIGHT OF THE TOTAL HEIGHT OF THE TOWER BE 10m!

• -21

aruntej pls explain the ans

• -14

This problem can be solved by using the nth second formula  i.e. , Snth = u + 1/2 a(2n-1)  . Snth is the distance travelled in nth second  , u=initial velocity  and  a = g

Let the total time taken  by the body to reach the bottom be n seconds  . So, the last two seconds will be n and (n-1).So, by putting the values   u=0 m/s and g=10m/s (as it is a freely falling body) we get two equations i.e.,
Snth = 0 + 1/2 g(2n-1)  = 5(2n-1)  .  ..........................1st equation
S(n-1)th = 0 + 1/2 g(2(n-1)-1)  =5(2n-3)  ..........................2nd equation
from 1 and 2 , given, 1+2 = 40 m.  So add both the equations
5(2n-1) +5(2n-3) =40  By solving this we get n=3 ( Hence total time ,t = 3 seconds)
By putting the values of u=0m/s ,g=10 m/s2 and t=3s  in the equation  S = ut + 1/2 gt2, We get Height of the Tower ,S =45 m .
If you are satisfied,then do give me thumbs up ,dude !!!

• 208

Let 'h' be the height ,'t' be the time of decent.

u=0, a= g = 10m/s2.

using:  s=ut+1/2 at2 . we get,

h=1/2 * 10 * t2

=> h=5t2 ------(1)

Since the body travels 40 m during last 2 sec.,Hence the body covers (h-40)m in time (t-2)sec

again using: s=ut+1/2at2  ,we get

h-40=1/2 * 10 (t-2)2

• 17

I think my solution is simpler than yours!!

• -21

(contd......)

h-10=5 (t-5)2 ------------(2)

(2)-(1)

40 = 5[t2-(t-2)2 ]

=> [t- t + 2] [t + t -2] = 8

=> 2t - 2 =4

on fuether simplification we get:

t=3 sec

from (1) we have

h=5t2

substituting value of t , we get

h = 5 * 3 * 3

=>h = 45 m

• 42

Intelligent guy

• -17
Please answer this question a body falls freely from a tower and travels a distance of tokin it's last two second . The height of the tower is

• 16
contd......)

h-10=5 (t-5)2?------------(2)

(2)-(1)

40 = 5[t2-(t-2)2?]

=> [t- t + 2] [t + t -2] = 8

=> 2t - 2 =4

on fuether simplification we get:

t=3 sec

from (1) we have

h=5t2

substituting value of t , we get

h = 5 * 3 * 3

=>h = 45 m
• -1
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