A rocket is launched straight up with constant acceleration. Four seconds after liftoff, a bolt falls off the rocket. The bolt hits the ground 8 seconds later. Three seconds after the bolt falls off, the rocket's engines fail. How high does the rocket go before it begins to fall back to earth?
[tex]x = x_0 + v_0 t + (1/2) a t^2[/tex]
[tex]v = v_0 + a t[/tex]
The Attempt at a Solution
I used the first equation to find the distance between launch and the bolt falling off. I got 313.6m
After that I used the values I had to find the acceleration until the engine failed. 39.2m/s^2 ?
Then I just calculated distances, velocities, and times with the values I had. Ultimately I found a maximum height of 3601.5m. For whatever reason this just doesn't seem right. At the moment the engine failed I found a height of 646m. For the rocket to travel that far with a negative acceleration doesn't seem kosher. Any help would be great. Thanks.