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} $
BETA
