Question

A particle is dropped from the top of a tower.It covers 40m in last 2 sec. Find the height of the tower

Answer 1

Answer: 45m

Explanation:

Let the body fall through the height of tower in t seconds. From,

D_n   =u+a/2  (2 n-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

so ans is 45m

Answer 2

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