Question

A ball is released from the top of a tower of height H metres. It takes T sec. to reach the ground.what is the position of the ball from ground in T/3 seconds.
(please tell the explianation)​

Answer 1

Answer:Let the acceleration be 'g'

Then initial velocity = 0 m/s [in T seconds]

Let the distance be 'h' m.

We know,

Equation of motion:

s = ut + 1/2 gT²

Then,

h = 1/2gT² ......(1)

h1 = 1/2gT²/9'......(2)

Substitute (1) and (2):

We'll get h1 = h9

∴ The distance from ground is h-h/9 = 8h/9 

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Explanation:

it takes t seconds to reach the ground . What is the position of the ball at t/3 seconds? distance,s -? so ans is, distance travelled is h/9 metres from tower.