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leaving the water vertically at 8.0 m/s, a steel head lands on the top of a...

leaving the water vertically at 8.0 m/s, a steel head lands on the top of a waterfall 1.8m high. how long is it in the air? (express your answer in seconds)

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Answer #1

Using S= ut + (1/2)at​2​​​​

We have, 1.8 = 8t - [(1/2)*9.8*t​​​​​​2​​​​​]

Solving this quadratic equation we get, t= 0.27 s or t= 1.36 s

The steelhead lands on the top of waterfall when it is decending so the time taken is 1.36 seconds.

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