Question

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 ratio 1:3, the smaller coming to rest. Find the distance from the launching point to the point where the heavier piece lands.

Answer 1

At the highest point$,$ the projectile has horizontal velocity$.$ The lighter part comes to rest$.$

Hence the heavier part will move with increased horizontal velocity$.$ in vertical direction$,$ both parts have zero velocity and undergo same acceleration$,$ hence they will cover equal vertical displacement in a given time$.$

thus$,$ both will hit the ground centre of ass$,$ the centre of mass hits the ground at the position where the origin projectile would have landed$.$

the range of the origin projectile is$:$

$X_{cm}=\frac{2u^2\sin \theta \cos \theta }{g}=\frac{2\times {10}^4\times \frac{3}{5}\times \frac{4}{5}}{10}=960m$

The centre of mass will hit the ground at this position$.$ 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 $X=480m.$ if the heavier block hits the ground at $x_2$

$X_{cm}=\frac{m_1{x}_1+m_2{x}_2}{m_1+m_2}$

$\Rightarrow 960=\frac{\frac{M}{4}\times 480+\frac{3M}{4}\times x_2}{M}$

$\Rightarrow x_2=1120m$

We get,

the distance from the launching point to the point where the heavier piece lands is $1120m.$