Verified Solution • 2025–26 Session moderate

QJEEM20Phf112: JEE Main 2020 Physics - Class 11 JEEM20Phf112 Solution: A ball is dropped from the top of a 100 m high tower on a…

A ball is dropped from the top of a 100 m high tower on a planet. In the last 12s{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 _____.

Step-by-Step Verified Solution

Let time to travel 81 m is t sec.

Time to travel 100 m is t + 12{1 \over 2} sec.

∴\therefore 81 = 12{1 \over 2} ×\times a ×\times t2

⇒\Rightarrow t = 92a9\sqrt {{2 \over a}}

And 100 = 12{1 \over 2} ×\times a ×\times (12+t)2{\left( {{1 \over 2} + t} \right)^2}

⇒\Rightarrow t+12t + {1 \over 2} = 102a10\sqrt {{2 \over a}}

⇒\Rightarrow 92a9\sqrt {{2 \over a}} + 12{1 \over 2} = 102a10\sqrt {{2 \over a}}

⇒\Rightarrow 2a\sqrt {{2 \over a}} = 12{1 \over 2}

⇒\Rightarrow a = 8 m/s2

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