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

Internal force do not effect the motion of the centre of mass, the centre of mass hits the ground at the position where the original projectile would have landed. The range of the original projectile is,

`X_("COM")=(2u^(2)sinthetacostheta)/g=(2xx10^(4)xx3/5xx4/5)/10` m
`=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 = 480 m. If the heavier block hits the ground at `x_(2)`, then
`x_("com")=(m_(1)x_(1)+m_(2)x_(2))/(m_(1)+m_(2))rArr960`
`=((m)(480)+(3m)(x_(2)))/((m+3m))x_(2)=1120m`