Question 60 of 308

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

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Answer: 8

$\dfrac{g{\left[t-\dfrac{1}{2}\right]}^{2}}{2}\times \dfrac{1}{2}=81$ ... (1)
$\dfrac{1}{2}g{t}^{2}=100$ …. (2)
devide equation \left(1\right)&\left(2\right)
$\dfrac{t}{t-\dfrac{1}{2}}=\dfrac{10}{9}$
$t=5s$
put t in equation $\left(2\right)$
$g=8\dfrac{m}{{s}^{2}}$