Q. In an experiment in physics class, a model solid-fueled rocket is fired vertically with an acceleration of 3 m/sec2 for 5 seconds. After that time, it's fuel is exhausted and it continues upwards as a free fall particle. (Take upwards to be the positive direction.) a) What is the maximum altitude the rocket reaches? Ans. x-x0=vo(t)+(1/2)a(t)^2 where x0=0 vo=0 a=3 m/sec^2 t=5 I get x= 37.5 meters v=0, a=-9.81m/s/s t=5 sec Using v = v0 + at, we get v0 = 49.05 m/s^2 While the rocket engine is firing a= 3. v0= 0 so v= 0+ 3(5)= 15 and you can use THAT as v0 after the engine shuts down. AFTER the rocket engine has shut off, the only acceleration is that of gravity: -9.81 so v= 15- 9.81t (t is now measured from engine cutoff, not initial launch). From that we get h= 15t- 4.9t2+ 122.5 (since the height was 37.5 m at engine cutoff, t=0). The maximum height occurs when v= 0: 15- 9.81t= 0 gives t= 15/9.81= 1.52 seconds. That height will be h= 15(1.52)- 4.9(1.52)^2+ 37.5= 48.97 m. b) What is the rocket's acceleration 6.5 seconds after the launch? Here is where I am stuck. Any ideas.... c) What is the total time of flight? Some ideas: Finally, the flight will be over when the rocket hits the ground: h=0. Solve 15t- 4.9t2+ 37.5= 0 to find the time AFTER CUTOFF that that occurs. You will need to add the inital 5 seconds to find the length of the entire flight. I tried to do this but my answer is wrong. ( The computer does not take it ) d) What is the rocket's total displacment during the entire flight? The answer is zero, but can anybody explain why?