# 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

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Let the initial velocity of firing the cannon shell be v m/s, its mass M kG, and the angle of projection with the horizontal be `alpha` .

Then the equations of motion are:

In the x-direction:

`v_x = vcosalpha`

`x(t) = vcosalpha * t`

In the y-direction:

`v_y = - g t +vsinalpha`

`y(t) = - 1/2 g t^2 + vsinalpha *t`

At the highest point `v_y` = 0,

`rArr t = (vsinalpha)/g` and

`x = (vcosalpha *vsinalpha)/g`

`=(v^2sin2alpha)/(2g)`

Total horizontal range=2x

`=(v^2sin2alpha)/g `

`=1/2` kM

`rArr x=1/4`

`y=-1/2*g*((vsinalpha)/g)^2+ vsinalpha * (vsinalpha)/g`

`=(v^2sin^2alpha)/(2g)`

**Explosion**: Horizontal momentum is conserved . Assuming no energy is released from the explosion itself,

`M *vcosalpha = 0+M/2 *V`

`rArr V = 2vcosalpha`

The equations of the second fragment, after the explosion, are:

`x(t) =2vcosalpha* t + 1/4`

`y(t) = -1/2 *g t^2 +(v^2sin^2alpha)/(2g)`

When it reaches the ground, y =0

So, `t =sqrt( (v^2sin^2alpha)/(g^2))`

`=(vsinalpha)/g`

`x = 2vcosalpha* (vsinalpha)/(g) + 1/4`

`=(v^2sin2alpha)/(g)+1/4`

=1/2+1/4 (since `(v^2sin2alpha)/(g)=1/2` kM)

=3/4 kM.

**Correct answer is at option D)**.