Question

it is from kinematics in vertical motion

Answer 1

Hello Friend,

Here

Initial height = h
Suppose the body falls for t seconds.

Body starts a free fall at t = 0
So, initial velocity u = 0

→ In t seconds,

s = ut + ½ at²
Here, h = 0 + ½ gt²
So, t² = 2h/g
So, t = √(2h/g) ----(1)

→ In (t-1) seconds,

s = ut + ½ a(t-1)²
So, s = ½ g (t-1)²

→ Now, if we subtract the distance travelled in (t-1) sec from the distance travelled in t sec, we can get the total distance travelled in the last second.

Here, distance travelled in last second = h-s = ½ gt² - ½ g (t-1)²

It is given in the question that it is equal to 11h/36

So,

½ gt²​ - ½g (t-1)² = 11h/36
So, ½g [ t² - (t-1)² ] = 11h/36
So, ½ g [ (t+t-1)(t-t+1) ] = 11h/36
So, ½ g [ (2t-1)(1) ] = 11h/36
So, g (2t-1) = 11h/18

So, (2t-1) = 11h/18g
So, 2t = (11h/18g) + 1
Squaring,
4t² = ((11h/18g) + 1)²
So, 4 (2h/g) = (11h +18g)² /18²g² (From 1)
So, 8 h/g × 18²g² = 121 h² + 396 hg + 324g²
So, 2592 hg = 121 h² + 396 gh + 324g²
Putting g = 10 m/s²
So, 25920 h = 121 h² + 3960 h +32400
So, 121 h² -21960h + 32400 = 0
So, 121h² - 21780h - 180h + 32400 = 0
So, 121h(h - 180) -180(h - 180) = 0
So, (h - 180)(121h - 180) = 0
So, h - 180 = 0 or 121h -180 = 0
So, h = 180 m or h = 180/121 m
So, h = 180 m or h ≈ 1.4876 m

Thus, here , h = 180 m

Hope it helps.

Purva
@Purvaparmar1405
Brainly.in