A body dropped from the top of a tower falls through 40m during the last two seconds of its fall. The height of tower is (g=10m/s2)
A
60m
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B
45m
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C
80m
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D
50m
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Solution
The correct option is B45m Let the body fall through the height of tower in t seconds.
We know that distance covered the body (Dn) at a particular second n is given by the relation Dn=u+a2(2n−1)
As the body is at rest initially ⇒u=0 ∴Dn=a2(2n−1)
Let t be the time taken by the body to reach ground.
Distance travelled in last 2 seconds =40m (given)
Total distance travelled in last 2 seconds of fall can also be expressed as D=Dt+D(t−1) =[0+g2(2t−1)]+[0+g2{2(t−1)−1}] =g2(2t−1)+g2(2t−3)=g2(4t−4) =102×4(t−1) ⇒40=20(t−1) t=2+1=3s
Distance travelled in t seconds is s=ut+12at2=0+12×10×32=45m