Kinematics Question 414

Question: A body dropped from top of a tower fall through 40 m during the last two seconds of it’s fall. The height of tower it’s $ (g=10m/s^{2}) $

Options:

A) 60 m

B) 45 m

C) 80 m

D) 50 m

Show Answer

Answer:

Correct Answer: B

Solution:

Let the body fall through the height of tower in t seconds. From, $ D _{n}=u+\frac{a}{2}( 2n-1 ) $ we have, total distance travelled in last 2 second of fall it’s $ D=D _{t}+{D _{( t-1 )}} $

$ =[ 0+\frac{g}{2}( 2t-1 ) ]+[ 0+\frac{g}{2}[ 2( t-1 ) ]-1 ] $

$ =\frac{g}{2}( 2t-1 )+\frac{g}{2}( 2t-3 )=\frac{g}{2}( 4t-4 ) $

$ =\frac{10}{2}\times 4( t-1 ) $

$ \text{or, 40=20}( t-1 )\text{ or t=2+1=3s} $ distance travelled at t second is $ \text{s=ut+}\frac{1}{2}at^{2}=0+\frac{1}{2}\times 10\times 3^{2}=45\text{ m} $



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