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 seconds? (a) 8h/9 meters from the ground (b) 7h/9 meters from the ground (c) h/9 meters from the ground (d) 7h/18 meters from the ground

Answer: (a). h = 1/2gT²; distance fallen in T/3: s = 1/2g(T/3)² = h/9 Height above ground = h - h/9 = 8h/9 So the answer is (a) 8h/9 m from the ground.

A ball is released from the top of a tower of height h meters. It… | JEE Main 2004 (AIEEE)

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Motion Under Gravity

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?

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Answer: (a)

$h = \frac{1}{2}gT^2$ ; distance fallen in $T/3$ : $s = \frac{1}{2}g\left(\frac{T}{3}\right)^2 = \frac{h}{9}$
Height above ground $= h - \frac{h}{9} = \frac{8h}{9}$
So the answer is (a) $\frac{8h}{9}$ m from the ground.

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