A hill is ` 500 m` hight. Supplies are to be across the hill using a canon that can hurl packetets at a speed of `125 m//s` pver the hill . The canon is located at a distance of ` 800 m` from the foot to hill and can be veoved 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 inwhich a pachet can reach on the ground across the hill ? Taje ` g= 10 m//s^2`.

Explanation

The initial minimum vertical velocity (u_y) required to just cross the hill is
` u_y gt- sqrt 2 gh gt- sqrt( 2 xx 10 xx 500) gt- 100 m//s.`
The initial horizontal velocity,
` u_x = sqrtu^2 -u_y^2 = sqrt ((125)^2 - (100)^2) = 75m//s`
Tiem talen to reach the top of the hill,
` t = sqrt ( 2h)/g = sqrt (2 xx 500)/(10)) = 10 s`
Horizontal distance covered in ` 10 s = u_x xx t`
`= 75 xx 10 = 750 m`
So, canon has to be moved through a dustance
`= 800 - 650 = 50 m`
Time taken to move the canon through 50 m oh ground `=(50)/(2) = 25 s`
Total time taken by packet to reach ground the hill `= 25+10+10= 45 s`