Answer:
Distances covered by a freely falling body during\[{{1}^{st}},\text{ }{{2}^{nd}},\text{ }{{3}^{rd}}...\text{ }{{n}^{th}}\] or its motion is. \[{{S}_{nth}}=U+\frac{a}{2}(2n-1)\] \[{{S}_{1st}}=\frac{9.8}{2}(2\times 1-1)\] \[{{S}_{2nd}}\text{ }=3\times 4.9m\] \[{{S}_{3rd}}=5\times 4.9m\] \[{{S}_{1}}:{{S}_{2}}:{{S}_{3}}=4.9:3\times 4.9:5\times 4.9=1:3:5\] \[{{S}_{1}}:{{S}_{2}}:{{S}_{3}}\]are proportional to odd number.
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