# Q. A cannon shell lands `1/2` km away from the cannon.A second shell fired identically,breaks into two equal parts at highest point.One part falls vertically.How far from the cannon will the other...

Q. A cannon shell lands `1/2` km away from the cannon.A second shell fired identically,breaks into two equal parts at highest point.One part falls vertically.How far from the cannon will the other land?

A) `1/2`

B) `1/3`

C) `2/3`

D) `3/4`

llltkl | Student

Let the initial velocity of the canonball be v and its angle of projection, `alpha` .

Since the time taken to return the components to the ground is equal to the time taken by the original projectile to reach the top of its flight, and the total flight of an unsplit identical projectile is 1/2km, the horizontal distance travelled by the second projectile before splitting is `1/2*1/2=1/4` km.

Time required to reach to the top of its flight=`1/4/(vcosalpha)`

The explosion leading to the split is instantaneous, and total horizontal momentum should be conserved.

Horizontal momentum before splitting`=Mvcosalpha` .

Given that one of the two equal fragments fall vertically from the point of splitting, its horizontal velocity must be zero.

Applying conservation of total horizontal momentum,

`Mvcosalpha=0+M/2*V` (where V is the horizontal component of velocity of the second fragment after the split)

`rArr V=2vcosalpha`

Horizontal distance travelled by the second projectile after splitting=velocity*time

`=2vcosalpha*1/4/(vcosalpha)=1/2` km

Therefore, the total horizontal distance travelled by the second projectile:

`=(1/2+1/4)=3/4` km (option D)