A rocket of mass 4.5 105 kg is in flight. Its thrust is directed at an angle of 47.5° above the horizontal and has a magnitude of 6.60 106 N. Find the magnitude and direction of the rockets acceleration. Give the direction as an angle above the horizontal.

Answer 1

Its acceleration will be in the direction of the net force

Force horizontal component= Fh = 6.6*10^6 cos 47.5

Force vertical component = Fv =
6.6*10^6 sin 47.5 - 4.5*10^5*9.81

so
Fh = 4.46*10^6 N
Fv = 2.19*10^6 N

F = sqrt(Fh^2+Fv^2) = 4.97*10^6 N

tan A = Fv/Fh = 2.19/4.46

Answer 2

now do A = F/m