Question

a body is released from the top of a tower of height H meters. it takes t time to reach ground. Where is the body t/2 time after the release​

Answer 1

☆$\red {HELLO\:DEAR}$☆

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

☆$\green {VISHU\:PANDAT}$☆

☆$\blue {FOLLOW\:ME}$☆

Answer 2

Answer:

3.at 3H/4 height from ground