Question

A ball is released from the top of the tower of height 'h'meters.it take t second to reach the ground.what is the position of the ball at t/3 second

Answer 1

8h / 9 meters

In this question we use the formula:

H = 1/2 x g x T^2 -------- Equation 1

Substituting the value of time in this formula we get:

S = 1/2 x g x (t/3)^2

S = 1/2 x g x T^2/9 ------- Equation 2

From equations 1 and 2 we get:

S = h/9

Now to calculate the position of the ball we use the formula:

S = h - h/9

Taking LCM we get:

S = 9 h - h / 9

S = 8 h / 9

Therefore, the position of the ball at 1/3 seconds is 8 h / 9.

Answer 2

Given,

➡️A ball is released from the top .

➡️Height = "h" meters

➡️time = "t" sec

To Find:-

⭐️The position of the ball at 1/3 second.

Calculations:-

H = 1/2 × g × T^2 -->(1)

Substituting values.

S = 1/2 × g × (t/3)^2

S = 1/2 × g × T^2/9 ---->(2)

From equation (1) and (2)

S = h/9

Position:-

S = h-h/9 (Take LCM)

S = 9h-h/9

S = 8h/9

The position of the ball at t/3 seconds is 8h/9.