Question #c022b

1 Answer
Aug 26, 2017

Answer:

#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.