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A body dropped from top of a tower fall through 40 m during the last two seconds of its fall. The height of tower is (g = 10 m/s 2 ) (a)60 m (b) 45 m (c)80 m (d) 50 m

A body dropped from top of a tower fall through 40 m during the last two seconds of its fall. The height of tower is (g = 10 m/s2)
(a)60 m
(b) 45 m
(c)80 m
(d) 50 m

Grade:10

3 Answers

Amit Saxena
35 Points
10 years ago

(b)
    Let the body fall through the height of tower in t seconds. From,
D_n   =u+a/2  (2n-1)  we have,total distance travelled in last 2 seconds of fall is
    D=  D_t+D_((t-1) )
    =   [ 0+ g/2 (2t-1)]+ [0+g/2 {2(t-1)-1}]
    =g/2  ( 2t-1)+g/2 (2t-3)=  g/2  (4t-4)
    =  10/2×4(t-1)
    Or, 40 = 20(t-1) or t =2+1 = 3s
    Distance travelled in t seconds is

s=ut +1/2   at^2  =0+1/2×10×3^2=45  m

teen adda
26 Points
6 years ago
Height. covered in t secondsh=1÷2gt^2Height covered in (t-2)secondsh`=1÷2g(t-2)^2Height covered in last two secondsh-h`=1÷2g[t^2 -(t-2)^2]40=1÷2×10(4t-4)t=3secPut this value for hh=1÷2×10(3)^2Therefore h=45m
Somnath Saha Chowdhury
13 Points
5 years ago
Let, The hight of the Tower =H m 
And the body takes T s to reach the ground.
So, H = 1/2 g T2  .....(1) [as body starts from rest so u=0]
Again , The body covered 40 m in last 2s. 
So we can write 
   (H  - 40)= 1/2 g(T - 2)2 
( 1/2 gT^2 - 40 )=1/2 g (T - 2)^2 (  From 1 H=1/2 gT^2)
T=3s
Puttin T =3s in 1 no equation , We get H = 45m .
 

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