Question #6ad4d

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Question #6ad4d

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Answer 1 · 2014-10-10T22:31:22

After t/2 seconds it has travelled a quarter of the total distance. This means that if the height of the tower is $h$ $h$ then it has fallen a distance of $\frac{h}{4}$ $h/4$ so is at a distance of $\frac{3 h}{4}$ $(3h)/4$ above the ground.

Explanation:

$h_{t} = \frac{1}{2} g . t^{2}$ $h_t=(1)/(2)g.t^2$

Where $h_{t}$ $h_t$ is the total height and g is the acceleration due to gravity.

After $\frac{t}{2}$ $(t)/2$ seconds :

$h_{\frac{t}{2}} = \frac{1}{2} {g (\frac{t}{2})}^{2}$ $h_(t/2)=(1)/(2)g((t)/(2))^2$

$= \frac{g (t^{2})}{8}$ $=g(t^2)/(8)$

So $h_{t} = 4 h_{t \frac{1}{2}}$ $h_t=4h_(t(1)/(2))$

So you can see that after half the time it has travelled a quarter of the distance.

This means that if the height of the tower is $h$ $h$ then it has fallen a distance of $\frac{h}{4}$ $h/4$ so is at a distance of $\frac{3 h}{4}$ $(3h)/4$ above the ground.

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Question #6ad4d

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