A ball is dropped from the top of a 100 m high tower on a planet. In the last $\frac{1}{2}\mathrm{s}$ before hitting the ground, it covers a distance of 19 m. Acceleration due to gravity (in ms^-2) near the surface on that planet is $\_\_\_\_\_\_$

For MO,

u = 0, x = 100, a = g′, t = t₀​

x = ut + 1/2at²

100 = 0×t + 1/2​ × g′ t₀²​

g′t₀² ​= 200    ...(1)

For MN

u = 0 , x = (100−19) = 81, a = g′ , t = (t₀​−1/2)

x = ut + 1/2​at²

81 = 0×t + 1/2​×g′×(t₀​−1/2​)²

81 = 2g′​(t₀​−1/2​)²

⇒g′(t₀​−1/2​)² = 162        ...(2)

In (1) and (2)

g′(t₀​−1/2​)²/g′t₀² ​​= 162/200​ = 81/100​

(t₀​−1/2)/​t₀ ​​= 09/10​

9t₀ ​= 10t₀ ​− 5

⇒g′ = ​200/(t₀²) ​= 200/25 ​= 8m/s²