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

The other half goes 17.4m high

Step-by-step explanation:

Pls see calculation in the attached file

Answer 2

`h = 83.093\,m`

Step-by-step explanation:

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

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

`v \approx 25.712\,(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\,(m)/(s)\right)^(2)+ 2 \cdot \left(9.807\,(m)/(s^(2))\right)\cdot (120\,m-65\,m)}`

`v_(o) \approx 32.845\,(m)/(s)`

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

`m \cdot (25.712\,(m)/(s) ) = 0.5\cdot m \cdot (32.845\,(m)/(s) ) + 0.5\cdot m \cdot v`

`25.842\,(m)/(s) = 16.423\,(m)/(s) + 0.5\cdot v`

`v = 18.838\,(m)/(s)`

The height reached by this half is:

`h = h_(o) -(v_(f)^(2)-v_(o)^(2))/(2\cdot g)`

`h = 65\,m - ((0\,(m)/(s) )^(2)- (18.838\,(m)/(s) )^(2))/(2\cdot (9.807\,(m)/(s^(2)) ))`

`h = 83.093\,m`