A) 60 m
B) 45 m
C) 80 m
D) 50 m
Correct Answer: B
Solution :
Let the body fall through the height of tower in t seconds. From, \[{{D}_{n}}=u+\frac{a}{2}\left( 2n-1 \right)\]we have, total distance travelled in last 2 second of fall is \[D={{D}_{t}}+{{D}_{\left( t-1 \right)}}\] \[=\left[ 0+\frac{g}{2}\left( 2t-1 \right) \right]+\left[ 0+\frac{g}{2}\left[ 2\left( t-1 \right) \right]-1 \right]\] \[=\frac{g}{2}\left( 2t-1 \right)+\frac{g}{2}\left( 2t-3 \right)=\frac{g}{2}\left( 4t-4 \right)\] \[=\frac{10}{2}\times 4\left( t-1 \right)\] \[\text{or, 40=20}\left( t-1 \right)\text{ or t=2+1=3s}\] Distance travelled at t second is \[\text{s=ut+}\frac{1}{2}a{{t}^{2}}=0+\frac{1}{2}\times 10\times {{3}^{2}}=45\text{ m}\]
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