Question

A ball is released from the top of a tower of height H meter.it takes T sec to reach the ground.what is position of the boll T/3 seconds

Answer 1

Answer:

The ball is positioned at a height of 8H/9 from the ground.

Explanation:

As the ball is released, initial velocity(u) = 0

  Given, S = H, t = T & a = g

Using,   S = ut + 1/2 at²

⇒ H = (0)T + 1/2 gT²

⇒ H = 1/2 gT²         ...(1)

At t = T/3,  let it at be distance 'x' from top

    Using, S = ut + 1/2 at²

⇒ x = (0)T/3 + 1/2 g(T/3)²

⇒ x = 1/2 gT²/9

⇒ x = H/9          [from (1)]

      therefore, it is a distance of H/9 from the top of the tower,  

whereas it is at a distance of H - H/9 from the ground.

The ball is positioned at a height of 8H/9 from the ground.

Answer 2

Explanation:

given :

• A ball is released from the top of a tower of height H meter.it takes T sec to reach the ground.what is position of the boll T/3 seconds

to find

• what is position of the boll T/3 seconds

let us do :

• 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)gT²

• h = (1/2)gT

• in T/3 sec

• h₁ = (1/2)gT² = (1/2) g. (T/3) = (1/2) T/9 ... (2)

• [1] and [2] we get h₁ = h/9 distance from

• point of release.

• therefore distance from ground is h. - h/9 = 8h/9