Answer:
16.4m while the pellet is going up, and 46.9m while the pellet is going down.
Explanation:
Taking the upward direction as the positive direction, the height of the balloon is [tex]y_b=y_{b0}+v_bt=(14 m)+(5.3 m/s)t[/tex] if we start counting t from the moment the pellet was fired ([tex]y_{b0}=14 m[/tex]), and since the balloon is moving at constant speed. The height of the pellet is [tex]y_p=y_{p0}+v_{p0}t+\frac{at^2}{2} =(38m/s)t+\frac{(-9.8m/s^2)t^2}{2}[/tex] since it's fired from the ground ([tex]y_{p0}=0m[/tex] ) and the acceleration of gravity pulls downwards. We want to know when [tex]y_b=y_p[/tex], that is, [tex]y_{b0}+v_bt=v_{p0}t+\frac{at^2}{2}[/tex], so we have to solve [tex]\frac{a}{2}t^2+(v_{p0}-v_b)t-y_{b0}=0[/tex], which has the form [tex]Ax^2+Bx+C=0[/tex] and we know must be solved using [tex]x=\frac{-B\pm\sqrt{B^2-4AC}}{2A}[/tex].
For our case, we have [tex](4.9m/s^2)t^2+(32.7m/s)t-14m=0[/tex], so our solution would be [tex]t=\frac{-32.7m/s\pm\sqrt{(32.7m/s)^2-4(4.9m/s^2)(14m)}}{2(4.9m/s^2)}[/tex], which gives the solutions [tex]t_+=6.21s[/tex] when using the + sign and [tex]t_-=0.46s[/tex] when using the - sign. This means that at t=0.46s the pellet is at the same height as the balloon while going up, then the pellet reaches maximum altitude and goes down, being again at the same altitude as the balloon at t=6.21s.
The easiest way to calculate the height of these points is calculating the height of the balloon at those times:
[tex]y_{b-}=y_{b0}+v_bt_-=14m+(5.3m/s)(0.46s)=16.44m[/tex]
[tex]y_{b+}=y_{b0}+v_bt_+=14m+(5.3m/s)(6.21s)=46.91m[/tex]
For verification, we could confirm this for the pellet:
[tex]y_{p-}=v_{p0}t_-+\frac{at_-^2}{2}=(38m/s)(0.46s)+\frac{(-9.8m/s^2)(0.46s)^2}{2}=16.44m[/tex]
[tex]y_{p+}=v_{p0}t_++\frac{at_-^2}{2}=(38m/s)(6.21s)+\frac{(-9.8m/s^2)(6.21s)^2}{2}=47.02m[/tex], this difference is due to rounding errors only, by using the exact value of [tex]t_+=6.21365241258s[/tex] one would get [tex]y_{b+}=y_{p+}=46.9323578m[/tex]
