Answer:
For the motion of
packet in the vertical direction
\[\upsilon _{y}^{2}\,-\upsilon
_{0y}^{2}=2gh\]\[\Rightarrow \,\upsilon _{y}^{2}-0=\,2\times 10\times 500\,\]
or \[{{\upsilon
}_{y}}=\,100\,m{{s}^{-1}}\]
\[\upsilon
=\,125\,m{{s}^{-1}}\]
\[\therefore \] \[{{\upsilon
}_{x}}=\,\sqrt{{{\upsilon }^{2}}-\upsilon _{y}^{2}}\,=\sqrt{{{(125)}^{2}}-\,{{(100)}^{2}}}\]
\[=75\text{ }m{{s}^{1}}~\]
Time of flight, \[T=\,\frac{2u\,\sin
\,\theta }{g}\,=\,\frac{2{{\upsilon }_{y}}}{g}\]
\[=\frac{2\,\times
\,100}{10}\,=20\,s.\]
Range of packet, \[R=\,{{\upsilon
}_{x}}\,T=\,75\,\times \,20=1500\,m\]
But actual range \[=800\text{
}m+800\text{ }m=1500\text{ }m\]
Therefore, canon
has to be moved towards the hill so that its distance from die foot of hill\[=750\text{
}m\].
Thus, to have range
= 1500 m, canon should move through a distance of 50 in.
Time taken by canon
to move through a distance of 50 m, \[t=\frac{50\,m}{2m{{s}^{-1}}}\,=25\,s.\]
\ Shortest time in which packet on
reach on the ground \[=25s+20s=45\text{ }s\].
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