Answer to Question #91270 in Mechanics | Relativity for Umar

Question

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]

Answer 1

(a) The rocket's velocity after 4.00 s

$$v_f=v_i+at=0+29.4\:\rm{m/s^2}\times 4.00\:\rm{s}=118\:\rm{m/s}$$

The rocket's position after 4.00 s

$$x_1=v_it+\frac{at^2}{2}=\frac{29.4\:\rm{m/s^2}\times (4.00\:\rm{s})^2}{2}=235\:\rm{m}$$

(b) The maximum height that rocket reaches

$$x_f=x_1+\frac{v_f^2}{2g}=235+\frac{118^2}{2\times 9.8}=945\:\rm{m}$$

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

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