I've been trying to figure this question out, and am now near on to pulling my hair out. A 1000 kg weather rocket is launched straight up. The rocket motor provides a constant acceleration for 16 seconds, then the motor stops. The rocket altitude 20 seconds after launch is 5100 meters. You can ignore any effects of air resistance. What was the rocket's acceleration during the first 16 seconds? In my panick for a solution, one thing I tried was this: v = u+at, = 0 +16a = 16a distance travelled during first 16 second s= ut +(1/2) at^2 = 0 + (1/2) at^2 = (1/2) a 20^2 = 200a total distance travelled after 20 second = 5100 m 5100 = 200a + 16a x4 -(1/2) x9.8 x4^2 264a = 5178.4 a = 19.61 m/sec^2 ....but obviously, this is incorrect. If anybody has any suggestions at all, I would be really grateful. Thanks!