Question 266: Gravitation

Question

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 _____.

Accepted Answer

Let time to travel 81 m is t sec.
Time to travel 100 m is t + $${1 \over 2}$$ sec.
$$ 	herefore $$ 81 = $${1 \over 2}$$ $$ 	imes $$ a $$ 	imes $$ t
2
$$ \Rightarrow $$ t = $$9\sqrt {{2 \over a}} $$
And 100 = $${1 \over 2}$$ $$ 	imes $$ a $$ 	imes $$ $${\left( {{1 \over 2} + t} ight)^2}$$
$$ \Rightarrow $$ $$t + {1 \over 2}$$ = $$10\sqrt {{2 \over a}} $$
$$ \Rightarrow $$ $$9\sqrt {{2 \over a}} $$ + $${1 \over 2}$$ = $$10\sqrt {{2 \over a}} $$
$$ \Rightarrow $$ $$\sqrt {{2 \over a}} $$ = $${1 \over 2}$$
$$ \Rightarrow $$ a = 8 m/s
2

Gravitation

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 _____.

Choose the correct answer: