A student launches a small rocket which starts from rest and from ground level (a height of h = 0). at a height of h = 1.95 km the rocket has reached a velocity of vf = 389 m/s. at that height the rocket runs out of fuel and therefore there is no longer any thrust propelling the rocket. you may assume that the positive direction is upwards in this problem.

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aristocles aristocles · Answer 1 · 2018-09-23T03:59:12+02:00

during the upward motion of rocket till its fuel is not over the acceleration is given by

[tex]v_f^2 - v_i^2 = 2 ad[/tex]

[tex]389^2 - 0 = 2 * a * 1950[/tex]

[tex]a = 38.8 m/s^2[/tex]

now the time taken by the rocket to reach that point is given as

[tex]v_f - v_i = a*t[/tex]

[tex]389 - 0 = 38.8 * t[/tex]

[tex]t = 10 s[/tex]

After fuel is over the maximum height that the rocket will reach is given by

[tex]v_f^2 - v_i^2 = 2 a d[/tex]

[tex]0^2 - 389^2 = -2*9.8*H[/tex]

[tex]H = 7720 m[/tex]

so the total height of rocket from ground is given by

[tex]H_{max} = 1.95 + 7.72 = 9.67 km[/tex]

now the time taken by the rocket to reach back on ground is

[tex] \Delta y = v_y*t + \frac{1}{2}gt^2[/tex]

[tex]-1950 = 389* t - 4.9 * t^2[/tex]

[tex] t = 84.11 s[/tex]

total time of motion of the rocket will be

T = 10 + 84.11 = 94.11 s