Answer:
a) [tex]h_m=74625\ m[/tex]
b) [tex]t=265.55\ s[/tex]
Explanation:
Given:
During the phase of fuel exhaustion:
velocity attained by the rocket just as the fuel ends:
[tex]v_f=u+a.t_f[/tex]
where:
[tex]u=[/tex] initial velocity of the rocket = 0
[tex]v_f=0+30\times 35[/tex]
[tex]v_f=1050\ m.s^{-1}[/tex] this will be the initial velocity for the phase of ascending of the rocket's height under the influence of gravity.
height at which the fuel finishes:
[tex]v_f^2=u^2+2a.h_f[/tex]
[tex]1050^2=0^2+2\times 30\times h_f[/tex]
[tex]h_f=18375\ m[/tex]
During the phase of ascend in height of rocket after the fuel is over:
Time taken to reach the top height after the fuel is over:
[tex]v=v_f+g.t'[/tex]
at top v = (final velocity during this course of motion )= 0 [tex]m.s^{-1}[/tex]
[tex]0=1050-9.8\times t'[/tex]
[tex]t'=107.1429\ s[/tex]
Height ascended by the rocket after the fuel is over:
[tex]v^2=v_f^2+2g.h'[/tex]
at the top height the velocity is zero
[tex]0^2=1050^2-2\times 9.8\times h'[/tex] (-ve sign denotes that the direction of motion is opposite to that of acceleration)
[tex]h'=56250\ m[/tex]
Therefore the maximum altitude attained by the rocket:
[tex]h_m=h_f+h'[/tex]
[tex]h_m=18375+56250[/tex]
[tex]h_m=74625\ m[/tex]
b)
time taken by the rocket to fall back to the earth:
[tex]h_m=v.t_m+\frac{1}{2} g.t_m^2[/tex]
where:
[tex]v=[/tex] initial velocity of the rocket during the course of free fall from the top height.
[tex]74625=0+4.9\times t_m^2[/tex]
[tex]t_m=123.41\ s[/tex]
Now the total time for which the rocket is in the air:
[tex]t=t_f+t'+t_m[/tex]
[tex]t=35+107.1429+123.41[/tex]
[tex]t=265.55\ s[/tex]
