Respuesta :
Answer:
Maximum height reached by the rocket is
[tex]y_{max} = 308 m[/tex]
total time of the motion of rocket is given as
[tex]T = 16.44 s[/tex]
Explanation:
Initial speed of the rocket is given as
[tex]v_i = 50 m/s[/tex]
acceleration of the rocket is given as
[tex]a = 2 m/s^2[/tex]
engine stops at height h = 150 m
so the final speed of the rocket at this height is given as
[tex]v_f^2 - v_i^2 = 2 a d[/tex]
[tex]v_f^2 - 50^2 = 2(2)(150)[/tex]
[tex]v_f = 55.68 m/s[/tex]
so maximum height reached by the rocket is given as the height where its final speed becomes zero
so we will have
[tex]v_f^2 - v_i^2 = 2 a d[/tex]
[tex]0 - 55.68^2 = 2(-9.81)(y - 150)[/tex]
[tex]y_{max} = 308 m[/tex]
Now the total time of the motion of rocket is given as
1) time to reach the height of 150 m
[tex]v_f - v_i = at[/tex]
[tex]55.68 - 50 = 2 t[/tex]
[tex]t_1 = 2.84 s[/tex]
2) time to reach ground from this height
[tex]\Delta y = v_y t + \frac{1}{2}gt^2[/tex]
[tex]-150 = 55.68 t - \frac{1}{2}(9.81) t^2[/tex]
[tex]t_2 = 13.6 s[/tex]
so total time of the motion of rocket is given as
[tex]T = 13.6 + 2.84 = 16.44 s[/tex]
