Question

ball is released from the top of a height h. It takes time T to reach the ground. Where is the ball at time T/2 from the ground?​

Answer 1

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+(1/2)gT

2

h=(1/2)gT

2

------[1]

in T/3 sec

h

1

=(1/2)gT

2

=(1/2)g(

3

T

)

2

=(1/2)g(

9

T

2

) -------[2]

from

[1] and [2] we get h

1

=

9

h

distance from point of release.

therefore distance from ground is h−

9

h

=

9

8h