The constant acceleration of a rocket launched upward, calculated knowing that the time it takes for a bolt that falls off the side of the rocker was 6.30 seconds, is 5.68 m/s².                                           Â
When the rocket is launched straight up with constant acceleration, the acceleration of the rocket is given by:
[tex] v_{f_{r}} = v_{i_{r}} + at [/tex] Â Â
Where: Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â
[tex]v_{f_{r}}[/tex]: is the final velocity of the rocket
[tex]v_{i_{r}}[/tex]: is the initial velocity of the rocket = 0
a: is the acceleration
t: is the time
After 4 seconds, the final speed of the rocket will be the initial speed of the bolt, so: Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â
[tex] v_{f_{r}} = v_{i_{b}} = at = 4a [/tex] Â
When the bolt falls off the side of the rocket, the bolt hits the ground 6.30 seconds later. Â Â Â Â
The initial height of the bolt will be the final height of the rocket, and vice-versa. With this, we can take the final height of the bolt as zero. Â Â Â Â Â Â Â Â Â Â Â Â
[tex] y_{f_{b}} = y_{i_{b}} + v_{i_{b}}t - \frac{1}{2}gt^{2} [/tex]
[tex] 0 = y_{i_{b}} + v_{i_{b}}t - \frac{1}{2}gt^{2} [/tex]
[tex] y_{i_{b}} = \frac{1}{2}9.81*(6.30)^{2} - 4a*6.30 = 194.7 - 25.2a [/tex]
Now, as we said above, this height (of the bolt) will be the final height of the rocket, so:
[tex] y_{f_{r}} = y_{i_{r}} + v_{i_{r}}t - \frac{1}{2}gt^{2} [/tex]
[tex] 194.7 - 25.2a = 0 + 0 - \frac{1}{2}a(4)^{2} [/tex] Â Â
[tex] a = \frac{194.7}{33.2} = 5.86 m/s^{2} [/tex] Â Â
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Therefore, the acceleration of the rocket is 5.68 m/s².
You can find another example of acceleration calculation here: https://brainly.com/question/24589208?referrer=searchResults
I hope it helps you! Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â
