Question

A projectile is fired from a gun at an angle of ${45}^0$ with the horizontal and with a speed of $20m/s$ relative to ground. At the highest point in its flight the projectile explodes into two fragments of equal mass. One fragment, whose initial speed is zero falls vertically. How far from the gun does the other fragment land, assuming a level terrain? take $g=10m/s^2$?

Answer 1

Suppose they don't explode then the centre of mass (whole projectile)

will fall on $A$

$R=\frac{u^2\sin 2\theta }{9}$

$=\frac{400\sin {90}^o}{9}$

$R=40m\Rightarrow OB=\frac{R}{2}=20m$

As the projectile get explodes due to internal forces

So the centre of mass will still fall on $A$

treating $O$ as origin

the coordinates of centre of mass

will be $xcm=\frac{m\times OB+m\times OC}{m+m}=\frac{20m+mx}{2m}$

but $xcm=OA=40$ meter

So

$40=\frac{20+x}{2}$

$\Rightarrow x=60$ meter