MHT CET 2021 Physics Solved Paper

(a) The distance travelled in nth second is given by
s=u+‌

1

2

a(2n−1)
s=‌

g

2

(2n−1)‌‌[‌ here ‌,u=0,a=g]
According to the question,
‌

11

36

h=‌

9.8

2

(2n−1)‌‌‌‌‌⋅⋅⋅⋅⋅⋅(i)
From second equation of motion,
h=‌

1

2

gn2
[∵u=0]...‌ (ii) ‌
From Eqs. (i) and (ii), we have
‌

11

36

×‌

9.8

2

×n2=‌

9.8

2

(2n−1) [here, g=9.8m∕s2 ]
⇒‌‌2n−1=‌

11

36

n2
⇒‌‌11n2−72n+36=0
⇒‌‌11n2−66n−6n+36=0
⇒‌‌11n(n−6)−6(n−6)=0
⇒‌‌n=6
(rejecting fractional values).
Therefore, h=‌

1

2

×10×6×6=180m