Book Online Demo

Check Your IQ

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

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 $\frac{T}{3} \sec$ ?

see full answer

Start JEE / NEET / Foundation preparation at rupees 99/day !!

21% of IItians & 23% of AIIMS delhi doctors are from Sri Chaitanya institute !!

An Intiative by Sri Chaitanya

\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.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

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}

Watch 3-min video & get full concept clarity

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Books for IIT JEE

Best Books for NEET

How to Join IIT after 10th

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Get Expert Academic Guidance – Connect with a Counselor Today!

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Books for IIT JEE

Best Books for NEET

How to Join IIT after 10th

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?

Detailed Solution

Best Courses for You

Get Expert Academic Guidance – Connect with a Counselor Today!

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 $\frac{T}{3} \sec$ ?