Physics - Center of Mass Question with Solution | TestHub

Let $h$ be the height above the ground at which the projectile explodes and $v_{jy}$ ($j=1,2,3$) the $y$-component of the velocity at the moment of the explosion for the three fragments. Label with 1 the fragment which lands after $t$ seconds. Since the landing time depends only on the $y$-component of the velocity at the moment of the explosion, we have $v_{3y}=v_{2y}$. We then need only the equations for fragments 1 and 2, stating that at landing time the $y$-coordinate will be equal to 0:

$$h+v_{1y}t-\frac{g{t}^2}{2}=0,h+v_{2y}(2t)-\frac{g{\left(2t\right)}^2}{2}=0$$

Since the masses of the three fragments are equal, and the $y$-component of the projectile's momentum at the moment of the explosion is equal to 0 (it explodes at the top of the trajectory), the momentum conservation law reads:

$$v_{1y}+v_{2y}+v_{3y}=0$$

The solution of the above equations is:

$$v_{1y}=-\frac{3}{4}gt,v_{2y}=v_{3y}=\frac{3}{8}gt$$

The height above the ground at which the projectile explodes is then:

$$h=\frac{5g{t}^2}{4}=\frac{5\times 10}{4}{\left(\frac{2}{5}\right)}^2=2\text{m}$$