Question #91270
A rocket moves straight upward, starting from rest with acceleration of 429.4m/s. It runs out of fuel at the end of 4.0s and continues to coast upward, reaching a maximum height before falling back to Earth.
a) find the rocket's velocity and position at the end of 4.00s. [v=118m/s, y=235m]
b) find the maximum height the rocket reaches. [945m]
c) find the velocity the instant before the rocket crashes on the ground [-136m/s]
[use a= -g= -9.80m/s square]
1
Expert's answer
2019-07-09T13:07:48-0400

(a) The rocket's velocity after 4.00 s

vf=vi+at=0+29.4m/s2×4.00s=118m/sv_f=v_i+at=0+29.4\:\rm{m/s^2}\times 4.00\:\rm{s}=118\:\rm{m/s}

The rocket's position after 4.00 s

x1=vit+at22=29.4m/s2×(4.00s)22=235mx_1=v_it+\frac{at^2}{2}=\frac{29.4\:\rm{m/s^2}\times (4.00\:\rm{s})^2}{2}=235\:\rm{m}

(b) The maximum height that rocket reaches

xf=x1+vf22g=235+11822×9.8=945mx_f=x_1+\frac{v_f^2}{2g}=235+\frac{118^2}{2\times 9.8}=945\:\rm{m}

(c) The velocity at the instant before the rocket crashes on the ground

v=2gxf=2×9.8×945=136m/sv=-\sqrt{2gx_f}=-\sqrt{2\times 9.8\times 945}=-136\:\rm{m/s}


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