Question

i want its complete solution

Answer 1

the acceleration of the ball will be g. Initial velocity will be 0 m/s in T sec. body travels h m.

By applying equations of motion we get

s= ut +1/2gT2

h = 1/2gT2 ------[1]

in T/4 sec 

so, height will be

h1 = (1/2).g.(T 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

thus,

height = (15/16)h

_____________________

related different question it's also important

A ball is released from the top of a tower of height h meters. It takes T second to reach the ground. What is the position f the ball in T/3 second –




As we know that ball is dropped from height H A and takes T seconds to fall on the ground, so it's initial velocity (u) will be 0.

So from 2 equation of motion we get, 

H=1/2gt2, as u=0........ (1)

Now let's consider the position of ballfrom air in T/3 seconds be H`. So again using 2 equation of motion, we get

H`=1/2gT2/9, as u=0.......(2)

Now from (1) and (2), we get

H`=H/9metres from the air.. 

So the location of ball from the ground will be=

H-H/9=9H-H/9=8H/9.

HENCE AFTER T/3 SECONDS POSITION OF BALL WILL BE 8H/9 METRES FROM THE GROUND.... 

Answer 2

$<b>●•••••••☆Hello User☆•••••••● 《《Here is your answer》》 <hr size = 3 width = 300 color = #000000 ><font color = #000000> ‍‌‌‌ Height of tower = h Time taken to reach to ground = T Now, by II equation of motion : s = ut + 1/2at² h = 0T + 1/2aT² a = 2h/T².......(1) Now . at time T/4 s = ut + 1/2at² s = 0T/4 + 1/2 × (2h/T²) × (T/4)² s = h/16.......from top of tower.. :. Position at T/4 is ( h - h/16 ) = ( 15h/16 ) from the ground. </font> <hr size = 3 width = 300 color = #000000 > ●•••••••☆Brainly Star☆•••••••●</b> <b>॥ॐ 卐 ☪ ✝ ॥</b>$