A ball is dropped from the top of a 100 m high tower on a planet. In the last ${1 \over 2}s$ 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 _____.
Answer (integer)
8
Solution
Let time to travel 81 m is t sec.
<br><br>Time to travel 100 m is t + ${1 \over 2}$ sec.
<br><br>$\therefore$ 81 = ${1 \over 2}$ $\times$ a $\times$ t<sup>2</sup>
<br><br>$\Rightarrow$ t = $9\sqrt {{2 \over a}}$
<br><br>And 100 = ${1 \over 2}$ $\times$ a $\times$ ${\left( {{1 \over 2} + t} \right)^2}$
<br><br>$\Rightarrow$ $t + {1 \over 2}$ = $10\sqrt {{2 \over a}}$
<br><br>$\Rightarrow$ $9\sqrt {{2 \over a}}$ + ${1 \over 2}$ = $10\sqrt {{2 \over a}}$
<br><br>$\Rightarrow$ $\sqrt {{2 \over a}}$ = ${1 \over 2}$
<br><br>$\Rightarrow$ a = 8 m/s<sup>2</sup>
About this question
Subject: Physics · Chapter: Gravitation · Topic: Gravitational Field and Potential
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