# Question c022b

Aug 26, 2017

$t = 1.010$ 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 $v t = 0$ 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.