Question

A firework shell is launched vertically upward from the ground with an initial speed of 44m/s. when the shell is 65 m high on the way up it explodes into two wequal mass halves, one half is observed to continue to rise straight up to a heigh of 120 m. How high does the other half go?

Answer 1

Answer:

The other half goes 17.4m high

Explanation:

Pls see calculation in the attached file

Answer 2

Answer:

$h = 83.093\,m$

Explanation:

The speed of the firework shell just before the explosion is:

$v = \sqrt{(44\,\frac{m}{s})^{2}-2\cdot \left(9.807\,\frac{m}{s^{2}}\right)\cdot (65\,m)}$

$v \approx 25.712\,\frac{m}{s}$

After the explosion, the initial speed of one of the mass halves is:

$v_{f}^{2} = v_{o}^{2} -2\cdot g \cdot s$

$v_{o}^{2} = v_{f}^{2} + 2\cdot g \cdot s$

$v_{o} = \sqrt{v_{f}^{2}+2\cdot g \cdot s}$

$v_{o} = \sqrt{\left(0\,\frac{m}{s}\right)^{2}+ 2 \cdot \left(9.807\,\frac{m}{s^{2}}\right)\cdot (120\,m-65\,m)}$

$v_{o} \approx 32.845\,\frac{m}{s}$

The initial speed of the other mass half is determined from the Principle of Momentum Conservation:

$m \cdot (25.712\,\frac{m}{s} ) = 0.5\cdot m \cdot (32.845\,\frac{m}{s} ) + 0.5\cdot m \cdot v$

$25.842\,\frac{m}{s} = 16.423\,\frac{m}{s} + 0.5\cdot v$

$v = 18.838\,\frac{m}{s}$

The height reached by this half is:

$h = h_{o} -\frac{v_{f}^{2}-v_{o}^{2}}{2\cdot g}$

$h = 65\,m - \frac{(0\,\frac{m}{s} )^{2}- (18.838\,\frac{m}{s} )^{2}}{2\cdot (9.807\,\frac{m}{s^{2}} )}$

$h = 83.093\,m$