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A ball is released from the top of a tower of height $' h' m$ . It takes $T seconds$ to reach the ground. What will be the position of the ball in $\frac{T}{3} \sec$ ?

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\frac{h}{9} m

from the ground

\frac{7 h}{9} m

from the ground

\frac{8 h}{9} m

from the ground

\frac{17 h}{9} m

from the ground

answer is C.

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Detailed Solution

A ball is released from the top of a tower of height

' h' m

. It takes

T seconds

to reach the ground. What is the position of the ball in

\frac{T}{3} \sec

\frac{8 h}{9} m

from the ground.
Given:
The acceleration of the ball will be

g

. Initial velocity

(u)

will be

0

. In

T seconds

the ball travels

' h' m

. From the second equation of motion, we have:

s = ut + \frac{1}{2} \times g t^{2}

, where

s

represents the position of the particle at any instant

t

.
But as the ball is falling down from a height

' h'

, the distance

s = h

= > h = \frac{1}{2} \times g T^{2} \dots \dots (1)

t = \frac{T}{3} s

let the height be

h_{1}

h_{1}

= \frac{1}{2} \times g t^{2}

= > h_{1} = \frac{1}{2} \times g {(\frac{T}{3})}^{2} \dots \dots . (2)

Comparing

(1)

and

(2)

, we get

h_{1}

=

\frac{h}{9}

Therefore the position of the ball from the ground,

h' = h - \frac{h}{9}

= > h' = \frac{8 h}{9}

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A ball is released from the top of a tower of height 'h' m. It takes T seconds to reach the ground. What will be the position of the ball in T3 sec?

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A ball is released from the top of a tower of height $' h' m$ . It takes $T seconds$ to reach the ground. What will be the position of the ball in $\frac{T}{3} \sec$ ?