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A cliff diver dove off a cliff at 8 m/s and landed in the water 24 m away from the cliff. How high is the cliff?

User Elvie
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1 Answer

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We are given the following information

Horizontal distance covered = 24 m

Initial speed = 8 m/s

We are asked to find the height of the cliff.

Let us first find the time.


\begin{gathered} v_(ix)=(\Delta x)/(t) \\ t=(\Delta x)/(v_(ix)) \end{gathered}

Where △x is the horizontal distance covered and vix is the initial horizontal speed of the diver.


t=(24)/(8)=3\; s

So, the time is 3 seconds.

The vertical distance covered by the diver is given by


y=(1)/(2)\cdot g\cdot t^2

Where g is the acceleration due to gravity that is 9.81 m/s^2


y=(1)/(2)\cdot9.81\cdot3^2=44.145\; m

Therefore, the height of the cliff is 44.145 m

User Dimitri Mostrey
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