Question

during your summer internship for an aerospace company, you are asked to design a small research rocket. the rocket is to be launched from rest from the earth’s surface and is to reach a maximum height of 970 m above the earth’s surface. the rocket’s engines give the rocket an upward acceleration of 16.0 m/s² during the time t that they fire. after the engines shut off, the rocket is in free fall. ignore air resistance. assume that the acceleration due to gravity does not change with the height of the rocket.

what must be the value of t in order for the rocket to reach the required altitude?

express your answer with the appropriate units.

Explanation:

Step 1: Analyze the motion phases

The rocket's motion has two phases: acceleration phase (with \(a = 16.0\ \text{m/s}^2\) for time \(T\)) and free - fall phase (with \(a=-g=- 9.8\ \text{m/s}^2\), where \(g = 9.8\ \text{m/s}^2\) is the acceleration due to gravity). Let the velocity at the end of the acceleration phase be \(v\) and the height reached during the acceleration phase be \(h_1\), and the height reached during the free - fall phase be \(h_2\). The total height \(H=h_1 + h_2=970\ \text{m}\).

For the acceleration phase:

We use the kinematic equations. The initial velocity \(u = 0\) (starts from rest).
The velocity at the end of the acceleration phase: \(v=u + aT=0 + 16T=16T\) (using \(v = u+at\))
The height reached during the acceleration phase: \(h_1=uT+\frac{1}{2}aT^{2}=0+\frac{1}{2}\times16\times T^{2}=8T^{2}\) (using \(s = ut+\frac{1}{2}at^{2}\))

For the free - fall phase:

At the maximum height, the final velocity \(v_f = 0\). We use the kinematic equation \(v_f^{2}-v^{2}=2ah_2\). Here, \(v_f = 0\), \(v = 16T\), and \(a=-g=-9.8\ \text{m/s}^2\)
\(0-(16T)^{2}=2\times(- 9.8)\times h_2\)
\(h_2=\frac{(16T)^{2}}{2\times9.8}=\frac{256T^{2}}{19.6}\)

Step 2: Set up the total height equation

The total height \(H = h_1+h_2\)
\(970=8T^{2}+\frac{256T^{2}}{19.6}\)
First, find a common denominator for the right - hand side. The common denominator of 1 and 19.6 is 19.6.
\(8T^{2}=\frac{8\times19.6T^{2}}{19.6}=\frac{156.8T^{2}}{19.6}\)
So, \(970=\frac{156.8T^{2}+256T^{2}}{19.6}=\frac{(156.8 + 256)T^{2}}{19.6}=\frac{412.8T^{2}}{19.6}\)
Then, solve for \(T^{2}\):
\(T^{2}=\frac{970\times19.6}{412.8}\)
\(T^{2}=\frac{19012}{412.8}\approx46.06\)
Take the square root of both sides:
\(T=\sqrt{46.06}\approx6.79\ \text{s}\)

Answer:

The value of \(T\) is approximately \(\boldsymbol{6.8\ \text{s}}\) (or more precisely, if we calculate \(\sqrt{\frac{970\times19.6}{412.8}}\):
First, \(970\times19.6 = 970\times(20 - 0.4)=970\times20-970\times0.4 = 19400 - 388 = 19012\)
\(\frac{19012}{412.8}\approx46.06\)
\(\sqrt{46.06}\approx6.79\ \text{s}\approx6.8\ \text{s}\))