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 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
$\frac{h}{9}$ meters from the ground
$\frac{7h}{9}$ meters from the ground
$\frac{8h}{9}$ meters from the ground
$\frac{17h}{18}$ meters from the ground
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
$\;\;h = ut + \frac{1}{2}g{t^2} \Rightarrow h = \frac{1}{2}g{T^2}$
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.$
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