A projectile is fired at a speed of $$100$$ ms−$$1$$ at an angle of $$370$$ 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

Question

A  projectile is fired at a speed of $$100$$ ms−$$1$$     at an angle of $$370$$     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

Infinity Learn · Accepted Answer

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. Thus, both will hit the ground together. As internal  forces do not affect 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.  $$xcm=2u2sin$$θ$$cos$$θ$$g=2$$×$$104$$×$$35$$×$$4510m=960$$ m     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 $$x2$$,     then $$xcm=m1x1+m2x2m1+m2$$       ⇒ $$960$$  $$=(m)(480)+(3m)(x2)(m+3m)$$     ⇒ $$x2=1120$$  m

A projectile is fired at a speed of 100 ms−1 at an angle of 370 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

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