Question

Ball A is released from top of tower when it descends by h1 ball b is released from h2 below the top of tower if both the ball Reach ground simultaneously find height of the tower​

Answer 1

Explanation:

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

Answer 2

HERE'S YOUR ANSWER BUDDY!!