Question

What is the distance covered by a free falling body during the first four seconds of its motion?​

Answer 1

Answer:

s(n)=g/2(2n-1)

put n=1

s(n)=10/2(2×1-1)

5×1m

5m...

hope it was helpful ..

Explanation:

Whether explicitly stated or not, the value of the acceleration in the kinematic equations is -9.8 m/s/s for any freely falling object. If an object is merely dropped (as opposed to being thrown) from an elevated height, then the initial velocity of the object is 0 m/s.

Answer 2

$\huge\mathcal{\fcolorbox{black}{black}{\blue{QUESTION࿐}}}$

What is the distance covered by a free falling body during the first four seconds of its motion?

$\huge\mathcal{\fcolorbox{black}{black}{\blue{ANSWER࿐}}}$

Initial velocity of the body u=0

Distance covered in n th

second, S n =u+ 21 g(2n−1)

∴ S n = 21

g(2n−1) (as u=0)

Distance moved in 1 st second i.e. n=1,

S 1 = 21 g(2×1−1)= 21 g

Distance moved in 2 nd second i.e. n=2,

S 2 = 21 g(2×2−1)= 23 g

Distance moved in 3 rd second i.e. n=3,

S 3 = 21 g(2×3−1)= 25 g

⟹ S 1 :S 2 :S 3 =1:3:5