question_answer 61)

                  A hill is 500 in high. Supplies are to be sent across die hill using a canon that can hurl packets at a speed of 125 in/s over the hill. The canon is located at a distance of 800 m from the foot of hill and can be moved on the ground at a speed of 2 m/s; so that its distance from the hill can be adjusted. What is the shortest time in which a packet can reach on the ground across the hill? Take \[g=10\text{ }m/{{s}^{2}}\].

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 \[=25s+20s=45\text{ }s\].