1. The problem statement, all variables and given/known data A rocket accelerates upwards from the ground at 25ms-2 for 2.5 s at an angle of 80 degrees to the horizontal. The rocket motor then stops and eventually falls to the ground. Neglecting air reststance and assuming thet the trajectory during acceleration is a straight line: a. Make a labelled sketch of the rockets trajectory starting from the launch point. Include the numerical values of the horizontal and vertical velocity at the point where the rocket motor stops. b. How long after launch does the rocket hit the ground? c. How far does the rocket travel horizontally from the launch point? 2. Relevant equations v=u2+2as S=ut+1/2at2 Uh=xcosθ Uv=xcosθ v=u+at and v=u+gt 3. The attempt at a solution See attached for diagram etc. I cannot actually upload it at this time, it will not lwt me? I can update it later if necessary. v=u+at =0+25*2,5 = 62.5m/s Uh = 62.5cos80 = 10.853 m/s Uv = 62.5sin80 = 61.55 m/s b. linear distance travelled under acceleration = s=0*t+1/2*25*2.52 = 78.125m This part is now for the point where there is no acceleration: Uh = 62.5cos80 = 10.853 m/s Uv = 62.5sin80 = 61.55 m/s Now V has changed to U. when V=0 at max height V2=U2+2S Thus S = -62.5/2*-9.81 = 3.19m extra to the top with no acceleration now. H1 = 78.125m + 3.19m = 81.315m For time to max height: V=U+gt = 0=-62.5*-9.81t Thus t2 = 6.37s Now time to drop from max height: -81.315 = 0*t+1/2*-9.81*t2 t=4s t1 = 2.5 sec = acceleration phase. t2 = 6.37 from after acceleration to the top t3 = 4 sec time to fall back down Total = 12.87 c. How far does the rocket travel: S=Uh*t Uh = 10.853 S=Uh*2.5= 27.1325 S=Uh*6.37= 69.13 S=Uh*4= 43.41 S = 139.67. I have done this and have no answers and do not know if it is right or where I might have gone wrong? Any help is much appreciated and thanks.