A projectile is fired at a speed of 100 m/s at an angle of $37^{o}$ above the horizontal. At the highest point, the projectile breaks into two parts of mass ration $1 : 3$ , the smaller coming to rest. Find the distance from the launching point to the point where the heavier piece lands.

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Solution

Internal forces do not affect the motion of the centre of mass, the centre of mass hits the ground at a position where the original projectile would have landed. The range of the original projectile is
$x_{C M} = \frac{2 u^{2} s i n θ c o s θ}{g} = \frac{2 \times 10^{4} \times \frac{3}{5} \times \frac{4}{5}}{10} m = 960 m$
The centre of mass will hit the ground at this point. As the smaller block comes to rest after breaking, it falls down vertically and hits the ground at half of the range, i.e., at $x = 480 m$ . If the heavier block hits the ground at $x_{2}$ , then
$x_{C M} = \frac{m_{1} x_{1} + m_{2} x_{2}}{m_{1} + m_{2}} ⟹= \frac{(m) (480) + (3 m) (x_{2})}{(m + 3 m)}$
$∴ x_{2} = 1120 m$

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A projectile is fired at a speed of 100 m/s at an angle of $37^{\circ}$ above the horizontal. At the highest point, the projectile breaks into two parts of mass in ratio 1 : 3, the smaller coming to rest. Find the distance from the launching point to the point where the heavier piece lands.