A body falling from rest under gravity passes a certain point $P$. It was at a distance of 400 m from $\mathrm{P}$, 4 $\mathrm{~s}$ prior to passing through P. If $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}$, then the height above the point P from where the body began to fall is

JEE Main · 2006 · 30 Apr 2006 · Offline · Chapter_pending

A body falling from rest under gravity passes a certain point

P

. It was at a distance of 400 m from

P

, 4

s

prior to passing through P. If

g = 10 m / s^{2}

, then the height above the point P from where the body began to fall is

Answer: (A)

Measure time from the instant the body is released, and let it reach

P

after

t

seconds, having fallen a height

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

. Four seconds earlier it was

400

m higher, so at time

t - 4

it had fallen

h - 400 = \frac{1}{2} g (t - 4)^{2}

. Subtracting the two,

400 = \frac{1}{2} g [t^{2} - (t - 4)^{2}] = \frac{1}{2} g (8 t - 16)

. With

g = 10

this is

400 = 5 (8 t - 16) = 40 t - 80

, which gives

t = 12

s. The height from which it began to fall is then

h = \frac{1}{2} \cdot 10 \cdot 1 2^{2} = 720

m above

P

JEE Main · 2006 · 30 Apr 2006 · Offline · Single correct option