Question

A ball is released from the top of a tower of height h meters. it takes T seconds to reach the ground. what is the position of the ball in T/3 second
a) h/9 meters from the ground.
b) 7h/9 meters from ground.
c) 8h/9 meters from the ground.
d) 17h/ meters from the ground.
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Answer 1

$\huge\blue{\underline {\overline{\boxed{ \mathbb{ \orange{: ANSWER :}}}}}}$

GIVEN :

Height of the tower = 'h' m.

Time taken by the ball to reach the ground = T sec.

TO FIND :

Position of the ball in $\sf \frac{T}{3}$ sec.

CORRECT OPTION :

$\sf\green{(c) \frac{8h}{9} \: meter\: from\:the \: ground.}$

SOLUTION :

Let the position of the ball in $\sf \frac{T}{3}$ sec be 'x' m from the ground.

Here, Initial velocity,u = O, Total Height = h, Total time = T and a = g.

We've, From 2nd Equation of motion

$\implies\sf{ \blue{s = ut + \frac{1}{2} {at}^{2} }}$

$\implies\sf{h = (0 \times T) + \frac{1}{2} { \times g \times T}^{2}}$

$\implies\sf{h = \frac{1}{2} { \times g \times T}^{2}}$

$\implies \: \sf{h = \frac{ {gT}^{2} }{2} }$

$\implies\sf{ {T}^{2} = \frac{2h}{g} } \: \: \: \: \: \: \: \: \: \: \: \: \rightarrow(1)$

Now, for travelling (h-x) m, Time required = $\sf \frac{T}{3}$ sec.

Again from 2nd Equation of motion we have,

$\sf{ \blue{s = ut + \frac{1}{2} {at}^{2} }}$

On putting the required values we get,

$\implies\sf{(h - x) = (0 \times T) + \frac{1}{2} { \times g \times { (\frac{T}{3} )}^{2}}}$

$\implies\sf{(h - x) =\frac{1}{2} { \times g \times { \frac{T}{9}}^{2}}}$

[Putting the value of ${T}^{2}$, from (1) we get]

$\implies\sf{(h - x) = \frac{1}{2} { \times g \times {\frac{ \frac{2h}{g} }{9}}}}$

$\implies\sf{(h - x) = \frac{h}{9} }$

$\implies\sf{ x= h - \frac{h}{9} }$

$\implies\sf{ x=\frac{8h}{9} }$

Hence, the position of the ball in $\sf \frac{T}{3}$sec. is $\sf \frac{8h}{9}$ m from the ground.

#answerwithquality #BAL

Answer 2

Answer:

GIVEN :

Height of the tower = 'h' m.

Time taken by the ball to reach the ground = T sec.

TO FIND :

Position of the ball in \sf \frac{T}{3}3T sec.

CORRECT OPTION :

\sf\green{(c) \frac{8h}{9} \: meter\: from\:the \: ground.}(c)98hmeterfromtheground.

SOLUTION :

Let the position of the ball in \sf \frac{T}{3}3T sec be 'x' m from the ground.

Here, Initial velocity,u = O, Total Height = h, Total time = T and a = g.

We've, From 2nd Equation of motion

\implies\sf{ \blue{s = ut + \frac{1}{2} {at}^{2} }}⟹s=ut+21at2

\implies\sf{h = (0 \times T) + \frac{1}{2} { \times g \times T}^{2}}⟹h=(0×T)+21×g×T2

\implies\sf{h = \frac{1}{2} { \times g \times T}^{2}}⟹h=21×g×T2

\implies \: \sf{h = \frac{ {gT}^{2} }{2} }⟹h=2gT2

\implies\sf{ {T}^{2} = \frac{2h}{g} } \: \: \: \: \: \: \: \: \: \: \: \: \rightarrow(1)⟹T2=g2h→(1)

Now, for travelling (h-x) m, Time required = \sf \frac{T}{3}3T sec.

Again from 2nd Equation of motion we have,

\sf{ \blue{s = ut + \frac{1}{2} {at}^{2} }}s=ut+21at2

On putting the required values we get,

\implies\sf{(h - x) = (0 \times T) + \frac{1}{2} { \times g \times { (\frac{T}{3} )}^{2}}}⟹(h−x)=(0×T)+21×g×(3T)2

\implies\sf{(h - x) =\frac{1}{2} { \times g \times { \frac{T}{9}}^{2}}}⟹(h−x)=21×g×9T2

[Putting the value of {T}^{2}T2 , from (1) we get]

\implies\sf{(h - x) = \frac{1}{2} { \times g \times {\frac{ \frac{2h}{g} }{9}}}}⟹(h−x)=21×g×9g2h

\implies\sf{(h - x) = \frac{h}{9} }⟹(h−x)=9h

\implies\sf{ x= h - \frac{h}{9} }⟹x=h−9h

\implies\sf{ x=\frac{8h}{9} }⟹x=98h

Hence, the position of the ball in \sf \frac{T}{3}3T sec. is \sf \frac{8h}{9}98h m from the ground