Answer:
[tex]u = 260.22m/s[/tex]
[tex]S_{max} = 1141.07ft[/tex]
Explanation:
Given
[tex]S_0 = 89.6ft[/tex] --- Initial altitude
[tex]S_{16.5} = 0ft[/tex] -- Altitude after 16.5 seconds
[tex]a = -g = -32.2ft/s^2[/tex] --- Acceleration (It is negative because it is an upward movement i.e. against gravity)
Solving (a): Final Speed of the rocket
To do this, we make use of:
[tex]S = ut + \frac{1}{2}at^2[/tex]
The final altitude after 16.5 seconds is represented as:
[tex]S_{16.5} = S_0 + ut + \frac{1}{2}at^2[/tex]
Substitute the following values:
[tex]S_0 = 89.6ft[/tex] [tex]S_{16.5} = 0ft[/tex] [tex]a = -g = -32.2ft/s^2[/tex] and [tex]t = 16.5[/tex]
So, we have:
[tex]0 = 89.6 + u * 16.5 - \frac{1}{2} * 32.2 * 16.5^2[/tex]
[tex]0 = 89.6 + u * 16.5 - \frac{1}{2} * 8766.45[/tex]
[tex]0 = 89.6 + 16.5u- 4383.225[/tex]
Collect Like Terms
[tex]16.5u = -89.6 +4383.225[/tex]
[tex]16.5u = 4293.625[/tex]
Make u the subject
[tex]u = \frac{4293.625}{16.5}[/tex]
[tex]u = 260.21969697[/tex]
[tex]u = 260.22m/s[/tex]
Solving (b): The maximum height attained
First, we calculate the time taken to attain the maximum height.
Using:
[tex]v=u + at[/tex]
At the maximum height:
[tex]v =0[/tex] --- The final velocity
[tex]u = 260.22m/s[/tex]
[tex]a = -g = -32.2ft/s^2[/tex]
So, we have:
[tex]0 = 260.22 - 32.2t[/tex]
Collect Like Terms
[tex]32.2t = 260.22[/tex]
Make t the subject
[tex]t = \frac{260.22}{ 32.2}[/tex]
[tex]t = 8.08s[/tex]
The maximum height is then calculated as:
[tex]S_{max} = S_0 + ut + \frac{1}{2}at^2[/tex]
This gives:
[tex]S_{max} = 89.6 + 260.22 * 8.08 - \frac{1}{2} * 32.2 * 8.08^2[/tex]
[tex]S_{max} = 89.6 + 260.22 * 8.08 - \frac{1}{2} * 2102.22[/tex]
[tex]S_{max} = 89.6 + 260.22 * 8.08 - 1051.11[/tex]
[tex]S_{max} = 1141.0676[/tex]
[tex]S_{max} = 1141.07ft[/tex]
Hence, the maximum height is 1141.07ft
