Answer:
5080.86m
Explanation:
We will divide the problem in parts 1 and 2, and write the equation of accelerated motion with those numbers, taking the upwards direction as positive. For the first part, we have:
[tex]y_1=y_{01}+v_{01}t+\frac{a_1t^2}{2}[/tex]
[tex]v_1=v_{01}+a_1t[/tex]
We must consider that it's launched from the ground ([tex]y_{01}=0m[/tex]) and from rest ([tex]v_{01}=0m/s[/tex]), with an upwards acceleration [tex]a_{1}=28m/s^2[/tex] that lasts a time t=9.7s.
We calculate then the height achieved in part 1:
[tex]y_1=(0m)+(0m/s)t+\frac{(28m/s^2)(9.7s)^2}{2}=1317.26m[/tex]
And the velocity achieved in part 1:
[tex]v_1=(0m/s)+(28m/s^2)(9.7s)=271.6m/s[/tex]
We do the same for part 2, but now we must consider that the initial height is the one achieved in part 1 ([tex]y_{02}=1317.26m[/tex]) and its initial velocity is the one achieved in part 1 ([tex]v_{02}=271.6m/s[/tex]), now in free fall, which means with a downwards acceleration [tex]a_{2}=-9,8m/s^2[/tex]. For the data we have it's faster to use the formula [tex]v_f^2=v_0^2+2ad[/tex], where d will be the displacement, or difference between maximum height and starting height of part 2, and the final velocity at maximum height we know must be 0m/s, so we have:
[tex]v_{02}^2+2a_2(y_2-y_{02})=v_2^2=0m/s[/tex]
Then, to get [tex]y_2[/tex], we do:
[tex]2a_2(y_2-y_{02})=-v_{02}^2[/tex]
[tex]y_2-y_{02}=-\frac{v_{02}^2}{2a_2}[/tex]
[tex]y_2=y_{02}-\frac{v_{02}^2}{2a_2}[/tex]
And we substitute the values:
[tex]y_2=y_{02}-\frac{v_{02}^2}{2a_2}=(1317.26m)-\frac{(271.6m/s)^2}{2(-9.8m/s^2)}=5080.86m[/tex]
