A1300 kg weather rocket accelerates upward at 10m/s2. it explodes 1.5 s after liftoff and breaks into two fragments, one twice as massive as the other. photos reveal that the lighter fragment traveled straight up and reached a maximum height of 500 m . part a what was the speed of the heavier fragment just after the explosion?

To solve this problem it is necessary to apply the conservation equations of the Momentum. The equation that represents such momentum conservation is given by,

Final momentum = Initial momentum

Where,

=mass light

= mass heavy

= velocity of mass light

= velocity of mass heavy

= final velocity

One of the masses is heavier than the other previously detached so we will assume that mass is 1/3 of the total, and the heavy mass is 2/3 of the total.

We start finding the initial velocity, then

The distance traveled is then calculated by the kinematic equations of motion,

The speed of the lighter piece is given also for the kinematic equation of movement,

Replacing at the conservation of momentum equation we have,

Therefore the speed of the heavier fragment just after the explosion is 18.52m/s against the direction of movement.

we have been given a quadratic function and we need to restrict the domain such that it becomes a one to one function.

we know that vertex of this quadratic function occurs at (5,2).

further, we know that range of this function is .

if we restrict the domain of this function to either or , it will become one to one function.

let us know find its inverse.

upon interchanging x and y, we get:

let us now solve this function for y.

hence, the inverse function would be if we restrict the domain of original function to and the inverse function would be if we restrict the domain to .