Question

a body is released from height h above the ground which takes t seconds the ground the position of the body after 1/2 second is​

Answer 1

$Sólútíóñ :-$

the acceleration of the ball will be g.

Initial velocity will be 0.

in T sec. body travels h mts.

by applying equations of motion we get

s= ut +1/2gT^2

h = 1/2gT2 ------[1]

in T/4 sec h1 = 1/2gT^2/16 -------[2]

from [1] and [2] we get h1 =h/16 distance from point of release.

therefore distance from ground is h-h/16

=15h/16