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2.Motion in Straight Line
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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 $\frac {T}{3}$ seconds
A
$\frac{h}{9}$ meters from the ground
B
$\frac{7h}{9}$ meters from the ground
C
$\frac{8h}{9}$ meters from the ground
D
$\frac{17h}{18}$ meters from the ground
(AIEEE-2004)
Solution

(c) $\;\;h = ut + \frac{1}{2}g{t^2} \Rightarrow h = \frac{1}{2}g{T^2}$
After $\frac{T}{3}$ seconds, the position of ball,
$h' = 0 + \frac{1}{2}g{\left( {\frac{T}{3}} \right)^2} = \frac{1}{2} \times \frac{g}{9} \times {T^2}$
$h' = \frac{1}{2} \times \frac{g}{9} \times {T^2}$$ = \frac{h}{9}\,m$ from top
Position of ball from ground $ = h – \frac{h}{9} = \frac{{8\;h}}{9}\,m.$
Standard 11
Physics
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