question_answer

What is the ratio of the distance travelled by a body falling freely from rest in the first, second, and third seconds of its fall.

Answer:

Here,\[\text{u }=0,\text{a}=\text{g}\]; Using the relation  \[{{\text{D}}_{\text{n}}}=\text{u }+\frac{a}{2}\left( \text{2n}-\text{1} \right)\] we have      \[{{\text{D}}_{\text{1}}}:{{\text{D}}_{\text{2}}}:{{\text{D}}_{\text{3}}}=\frac{g}{2}\left( \text{2}\times \text{1}-\text{1} \right)\text{ }:\frac{g}{2}\left( \text{2}\times \text{2}-\text{1} \right):\frac{g}{2}\left( \text{2}\times \text{3}-\text{1} \right)\text{ }=\text{ 1}:\text{3}:\text{5}\]