A) \[1:4:9\]
B) \[1:2:4\]
C) \[1:3:5\]
D) \[1:2:3\]
Correct Answer: C
Solution :
Using the relation \[s=ut+\frac{1}{2}g{{t}^{2}}\] As the body is falling from rest, \[u=0\] \[s=\frac{1}{2}g{{t}^{2}}\] Suppose the distance travelled in \[t=2s,\text{ }t=4s,\text{ }t=6s\] are\[{{s}_{2}},{{s}_{4}}\]and\[{{s}_{6}}\]respectively Now, \[{{s}_{2}}=\frac{1}{2}g{{(2)}^{2}}=2g\] \[{{s}_{4}}=\frac{1}{2}g{{(4)}^{2}}=8g\] \[{{s}_{6}}=\frac{1}{2}g{{(6)}^{2}}=18g\] Hence, the distance travelled in first two seconds \[{{({{s}_{i}})}_{2}}={{s}_{2}}-{{s}_{0}}\] \[=2g\] \[{{({{s}_{m}})}_{2}}={{s}_{4}}-{{s}_{2}}\] \[=8g-2g=6g\] \[{{({{s}_{f}})}_{2}}={{s}_{6}}-{{s}_{4}}=18g-8g=10g\] Now, the ratio becomes \[=2g:6g:10g=1:3:5\]
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