Answer:
[tex]h = 83.093\,m[/tex]
Explanation:
The speed of the firework shell just before the explosion is:
[tex]v = \sqrt{(44\,\frac{m}{s})^{2}-2\cdot \left(9.807\,\frac{m}{s^{2}}\right)\cdot (65\,m)}[/tex]
[tex]v \approx 25.712\,\frac{m}{s}[/tex]
After the explosion, the initial speed of one of the mass halves is:
[tex]v_{f}^{2} = v_{o}^{2} -2\cdot g \cdot s[/tex]
[tex]v_{o}^{2} = v_{f}^{2} + 2\cdot g \cdot s[/tex]
[tex]v_{o} = \sqrt{v_{f}^{2}+2\cdot g \cdot s}[/tex]
[tex]v_{o} = \sqrt{\left(0\,\frac{m}{s}\right)^{2}+ 2 \cdot \left(9.807\,\frac{m}{s^{2}}\right)\cdot (120\,m-65\,m)}[/tex]
[tex]v_{o} \approx 32.845\,\frac{m}{s}[/tex]
The initial speed of the other mass half is determined from the Principle of Momentum Conservation:
[tex]m \cdot (25.712\,\frac{m}{s} ) = 0.5\cdot m \cdot (32.845\,\frac{m}{s} ) + 0.5\cdot m \cdot v[/tex]
[tex]25.842\,\frac{m}{s} = 16.423\,\frac{m}{s} + 0.5\cdot v[/tex]
[tex]v = 18.838\,\frac{m}{s}[/tex]
The height reached by this half is:
[tex]h = h_{o} -\frac{v_{f}^{2}-v_{o}^{2}}{2\cdot g}[/tex]
[tex]h = 65\,m - \frac{(0\,\frac{m}{s} )^{2}- (18.838\,\frac{m}{s} )^{2}}{2\cdot (9.807\,\frac{m}{s^{2}} )}[/tex]
[tex]h = 83.093\,m[/tex]
