A projectile is fired at a speed of 100 m/s at an angel of `37^0` 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 where the heavier piece lands.

Explanation

See ure. At the highest point, the projectrile has horizontal velocity. The lighter part comes to rest. Hence the heavier part willl move with inceased horizontal velocity. In vertical direction both parts have zero velocity and undergo same acceleration, hence they willl cover equal vertical displacements in a given time. Thus both will hit the ground together As interN/Al forces do not affect the motion of teh centre of mass, the centre of mass hits the ground at teh positin where the origiN/Al projectile would have landed. Te range of the origiN/Al projectile is
`x_(CM)=(2u^2sinthetacostheta)/g=(2xx10^4xx3/5xx4/5)/10m`
=960m.
The centre of mass will hit the grouond at this position. As the smaller block comes to rest after breaking. It falls down vertically and hits the ground at half othe range i.e., at `x=480m`. If the heavier block hits the ground at `x_2` then
`x_(CM)=(m_1x_1+m_2x_2)/(m_2+m_2)`
`960m=(M/4xx480m+(3M)/4xxx_2)`
or,
`x_2=1120m`