JEE MAIN - Physics (2004 - No. 68)
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?
$${{8h} \over 9}$$ meters from the ground
$${{7h} \over 9}$$ meters from the ground
$${h \over 9}$$ meters from the ground
$${{7h} \over {18}}$$ meters from the ground
Explanation
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.
$$\therefore$$ $$h = {1 \over 2}g{T^2}$$
After $$T/3$$ second, vertical distance moved by the ball
$$h' = {1 \over 2}g{\left( {{T \over 3}} \right)^2} $$
$$\Rightarrow h' = {1 \over 2} \times {{8{T^2}} \over 9}$$
$$ = {h \over 9}$$
$$\therefore$$ Height from ground
$$ = h - {h \over 9} = {{8h} \over 9}$$
Initial speed of ball is zero and it take T second to reach the ground.
$$\therefore$$ $$h = {1 \over 2}g{T^2}$$
After $$T/3$$ second, vertical distance moved by the ball
$$h' = {1 \over 2}g{\left( {{T \over 3}} \right)^2} $$
$$\Rightarrow h' = {1 \over 2} \times {{8{T^2}} \over 9}$$
$$ = {h \over 9}$$
$$\therefore$$ Height from ground
$$ = h - {h \over 9} = {{8h} \over 9}$$
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