A cannonball of mass 29.3 kg is shot into the air at an angle of 45 degrees with respect to horizontal. At the top of its ballistic trajectory, the cannonball suddenly explodes into two separate pieces, a smaller piece this is 1/3 the original mass, and a larger piece that is 2/3 the original mass. The smaller piece ends up landing right back at the spot where the cannonball was originally launched, 12.0 seconds after it was launched. How far from the launch site does the larger piece travel? You can neglect the effects of air resistance. (Hint: to use momentum in this problem, you'll want to go back and review your ballistic motion equations from week 2)

The distance from the launch site is x

Explanation:

In this question we are provided with some details which include

  The mass of the Cannonball M = 29.3 kg

   The large piece of the ball  

  The  mass of the small piece of the  ball

From motion on a projectile, mathematically the time to reach the maximum height

                  given that g =9.8 m/s

The 29.8 is divided by 2 because it is time taken for the journey of the cannonball

    =>  

The maximum height  

Now after the explosion

      Let   have a velocity of    

      And   have a velocity of

We are able to calculate for the small piece of the cannonball because it took the path of the main cannonball

Considering the law of conservation of momentum

Mathematically the expression for the range

Now to obtain the distance of the larger part from the original source of projection

       on landing

      We make use of this projectile expression

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volts times amps 6 x 0.5 3w