Question

A Stone is dropped from a bejght of 125 If g =10m^-2
What is the ration of the distances travelled by it during
the first and the last second of its mation​

Answer 1

Question:-

stone is dropped from a height of 125 m. If g = 10 m s $\large ^{-2}$ , what is the ratio of the distances travelled by it during the first and the last second of its motion?

Answer : 1:9

Formula used: v = u + at

Step by Step Explanation:

In v = u + at,

v = final velocity

u = Initial Velocity

a = acceleration

t = time

Now, $\large S_{1} = \frac {1}{2} \times g \times 1 = 5 m$

$\large t = \sqrt{ \frac{ 2h }{ g } } = \sqrt{ \frac{ 2 \times 125 }{ 10 } } = 5s$

$\large S_{5} = \frac {g}{2} (2(5) – 1)$

$\large = 9 \times 5 = 45$

$\large S_{1} : S_{5} = 1 : 9$