Question 94: 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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