Question

A rocket is launched straight up with constant acceleration. Four seconds after liftoff, a bolt falls off the side of the rocket. The bolt hits the ground 6.30 seconds later. What is the acceleration of the rocket?

Answer 1

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.                                                                                      

When the rocket is launched straight up with constant acceleration, the acceleration of the rocket is given by:

$v_{f_{r}} = v_{i_{r}} + at$    

Where:                                                              

$v_{f_{r}}$: is the final velocity of the rocket

$v_{i_{r}}$: 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:                                              

$v_{f_{r}} = v_{i_{b}} = at = 4a$  

When the bolt falls off the side of the rocket, the bolt hits the ground 6.30 seconds later.        

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.                        

$y_{f_{b}} = y_{i_{b}} + v_{i_{b}}t - \frac{1}{2}gt^{2}$

$0 = y_{i_{b}} + v_{i_{b}}t - \frac{1}{2}gt^{2}$

$y_{i_{b}} = \frac{1}{2}9.81*(6.30)^{2} - 4a*6.30 = 194.7 - 25.2a$

Now, as we said above, this height (of the bolt) will be the final height of the rocket, so:

$y_{f_{r}} = y_{i_{r}} + v_{i_{r}}t - \frac{1}{2}gt^{2}$

$194.7 - 25.2a = 0 + 0 - \frac{1}{2}a(4)^{2}$    

$a = \frac{194.7}{33.2} = 5.86 m/s^{2}$    

   

Therefore, the acceleration of the rocket is 5.68 m/s².

You can find another example of acceleration calculation here: brainly.com/question/24589208?referrer=searchResults

I hope it helps you!