Answer:
The time interval the rocket is in motion above the ground is the time in the two times the motion is going on [tex]T_{total} = 23,14 s[/tex]
Explanation:
[tex]V_{i} = 80 (\frac{m}{s} )[/tex]
[tex]S_{i} = 0 m[/tex]
The motion in the first step has an acceleration [tex]a_{1} = 4 (\frac{m}{s^{2} } )[/tex]
and the maximum height will be and the end of this step is [tex]S_{1} = 1000 m[/tex]
So to know the time until the rocket fail and change the acceleration:
[tex]S_{1} = S_{i} + V_{i1} * t + \frac{1}{2} * a_{1} *t^{2}[/tex]
[tex]1000m = 0 m + 80 \frac{m}{s} * t + \frac{1}{2} * 4 \frac{m}{s^{2} } * t^{2}[/tex]
[tex]1000= 2*t^{2} + 80 * t[/tex] you can divide the expression by two and simplify the calculating
[tex]2t^{2} + 80*t -1000=0[/tex]
[tex]t^{2} + 40*t -500=0[/tex]
Using quadratic equation :
[tex]\frac{-b +/- \sqrt{b^{2}-4*a*c } }{2*a}[/tex]
[tex]\frac{- 40+/- \sqrt{40^{2}-4*-500 } }{2}[/tex]
[tex]-20 +/- 30[/tex]
[tex]x_{1}= -50 , x_{2}= 10 ,[/tex] The time can be negative so, the time we are going to use is 10s
[tex]t_{1}= 10 s[/tex]
Now when the rocket fail it change the direction of the motion and the time is going to be the time it takes to reach earth again
[tex]v_{f} = v_{i}+a_{1}*t_{1}[/tex]
[tex]v_{f}= 80 \frac{m}{s} + 4 \frac{m}{s^{2} } * 10 s = 120 \frac{m}{s}[/tex]
[tex]S_{2} = S_{1} + V_{i2} * t + \frac{1}{2} * a_{2} *t^{2}[/tex]
[tex]0m = 1000m + 120 \frac{m}{s} * t +\frac{1}{2} (-29,8 \frac{m}{s^{2} })*t^{2}[/tex]
[tex]0m = 1000m + 120 \frac{m}{s} * t -14,9 \frac{m}{s^{2} })*t^{2}[/tex]
[tex]\frac{-b +/- \sqrt{b^{2}-4*a*c } }{2*a}[/tex]
[tex]\frac{- 120+/- \sqrt{120^{2}-4*-14,9*1000 } }{2*14,9}[/tex]
[tex]x_{1}= -5,108s , x_{2}= 13,14 ,[/tex] The time can be negative so, the time we are going to use is 13,14s
So the full time is the both times adding them
[tex]T_{total}= 10 + 13,14 = 23,14 s[/tex]
