Question

A freely falling particle covers a building of 45m height
in one second. Find the height of the point from where
the particle was released.​

Answer 1

Answer:

Sn = ut + g(2n-1)/2

45 = 0 + 10(2n-1)/2

90 = 10(2n-1)

2n-1 = 9

n = 5

therefore the particle covers 45 meters in fifth second of its journey

Now,

s=ut+at^2/2

s = at^2/2

s = 10×25/2

s = 125 meters (approximate answer because I took g = 10m/s^2)

Therefore the particle was thrown from a height of 125 meters which is the relative height of the two points.

hope this helps you

Answer 2

Answer:

if a particle covers 45m h in one sec..

sth=ut+g(2n-1)/2

45= 0t+10(2n-1)/2

9=2n-1

5=n

now,

s=ut+(gt^2)/2

s= 0t+ (10×5×5)/2

s= 250/2

s=125m

hope you understand!!