A ball is released from top of a tower of height h meters. It takes T, seconds to reach ground. posi

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2.Motion in Straight Line

medium

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 at $\frac{T}{3}\, second$

$\frac{{8h}}{9}$ $meters$ from the ground

$\frac{{7h}}{9}$ $meters$ from the ground

$\frac{h}{9}$ $meters$ from the ground

$\frac{{17h}}{{18}}$ $meters$ from the ground

(AIIMS-2012)

Solution

$\begin{array}{l}
h = \frac{1}{2}g{T^2}\\
now\,for\,t = T/3\,{\rm{second}}\,vertical\,{\rm{distance}}\,\\
moved\,is\,given\,by\\
h' = \frac{1}{2}g{\left( {\frac{T}{3}} \right)^2} \Rightarrow h' = \frac{1}{2} \times \frac{{g{T^2}}}{9} = \frac{h}{9}\\
\therefore \,Position\,of\,ball\,from\,ground\, = h – \frac{h}{9}\\
= \frac{{8h}}{9}
\end{array}$

Standard 11

Physics

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easy

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 at $\frac{T}{3}\, second$

Solution

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