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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/s²)
(a)60 m
(b) 45 m
(c)80 m
(d) 50 m

Hrishant Goswami , 11 Years ago

Grade 10

3 Answers

Amit Saxena

Last Activity: 11 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

Last Activity: 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

Last Activity: 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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