A ball is dropped form 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 $\mathrm{ms}^{-2}$) near the surface on that planet is ____.

JEE Main · 2020 · 08 Jan 2020 · Shift 2 · Kinematics

A ball is dropped form the top of a 100 m high tower on a planet. In the last

\frac{1}{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: 8

Solution

Let the fall take

t

seconds, so the full height gives

\frac{1}{2} g t^{2} = 100

. The last half second covers

19

m, so the first

(t - 0.5)

s cover

100 - 19 = 81

m, i.e.

\frac{1}{2} g (t - 0.5)^{2} = 81

. Dividing the two,

\frac{t ^{2}}{( t - 0.5 ) ^{2}} = \frac{100}{81}

, so

\frac{t}{t - 0.5} = \frac{10}{9}

, giving

t = 5

s. Then

g = \frac{200}{t ^{2}} = \frac{200}{25} = 8 m/ s^{2}

JEE Main · 2020 · 08 Jan 2020 · Shift 2 · Integer answer

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