Question

Question #c022b

Answer 1

`t = 1.010` s.

Explanation:

There’s a little bit of ambiguity in this question: does it mean to rise up 5 m and fall back down or just to rise 5 m? I’ll write a solution for both as both solutions are just applications of constant acceleration equations and the former is simply the latter solution multiplied by two.

Time to rise up to 5 m.
Write down what you know:
`s = 5` m
`u =`?
`v = 0` m s⁻¹ (it rises up until it slows down to a halt before falling back down.
`a = -9.8` m s⁻²
`t =` ?

Use this equation: `s = vt - ½at^2`
Note that `vt = 0` because `v=0`.

Make `t` the subject: `⇒ t = sqrt ((-2s)/a)`

Substitute in the values: `⇒ t = ± sqrt ((-2 × 5)/-9.8) = ± 1.010` s.

Obviously the time cannot be negative so the solution for this problem is `t = 1.01` s.

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If the question requires the time to rise 5 m and fall back down then double the time ⇒ `t = 2.02` s.