Question 97: Motion in 1D

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 ${T \over 3}$ seconds?

Accepted Answer

We know that equation of motion, $s = ut + {1 \over 2}g{t^2},\,\,$
Initial speed of ball is zero and it take T second to reach the ground.
$	herefore$ $h = {1 \over 2}g{T^2}$
After $T/3$ second, vertical distance moved by the ball
$h' = {1 \over 2}g{\left( {{T \over 3}} ight)^2} $
$\Rightarrow h' = {1 \over 2} 	imes {{8{T^2}} \over 9}$
$ = {h \over 9}$
$	herefore$ Height from ground
$ = h - {h \over 9} = {{8h} \over 9}$

Answer

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

Answer

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

Answer

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

Answer

${{7h} \over {18}}$ meters from the ground

Motion in 1D

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 \over 3} seconds?

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