A bottle rocket takes off with a = 34.5 m/s ^ 2 It fires until it reaches a height of 200 m, then shuts off, and gravity takes over. How much time does it take to reach its maximum height ?

Question

You need to find the second

Answer 1

Hi Aml F.,

The time for the rocket to reach maximum height will be in two stages, stage one for when it is accelerated upward, and stage two for when the rocket is in free fall to its maximum height.

For the time the rocket is accelerated at, a = 34.5 m/s², to a height of, y = 200 m, we will use y = y₀ + v₀t + .5at², where y₀ = 0 and v₀ = 0, we have y = .5at² or t = ±√(2*y/a) and we choose (+) time:

t₁ = √(2*200/32.5), (t₁ = time for stage one)

t₁ = 3.4 s, (the rocket fires and accelerated upward for this time)

We can find the velocity at time t₁, using, v = v₀ + at, where, v₀ = 0, therefore v = at:

v = at = 34.5*3.4

v = 117.5 m/s, (this v is the initial velocity for stage two free fall)

To find the time for stage two of the rocket in free fall, we can use v = v₀ + at, where v₀ = 117.5 m/s and v = 0 (the velocity at maximum):

v = v₀ + at

0 = 117.5 - 9.8t₂, (a = g = -9.8 m/s² and t₂ = stage two free fall)

t₂ = 117.5/9.8

t₂ = 12.0 s

The total time to reach maximum is t₁ + t₂ = 3.4 + 12.0 = 15.4 s.

I hope this helps, Joe.