Question

A hill is 500 m high. Supplies are to be sent across the hill using a canon that can hurl packets at a speed of 125 m/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 the 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 $m/s^{2$.

Answer 1

Answer:

Explanation:

According to the problem, speed of packets =125 m/s , height of the hill = 500 m, distance between the cannon and the foot of the hill, d = 800 m .

To cross the hill in shortest time, then the vertical component of the velocity should be minimum so that it just crosses the height of hill.

Distance through which canon has to be moved = 800 – 750 = 50 m Speed with which canon can move = 2 m/s

Answer 2

Answer:

hope you understand this

Explanation:

Solution :

The initial minimum vertical velocity (u_y) required to just cross the hill is

uy>−2–√gh>−2×10×500−−−−−−−−−−√>−100m/s.

The initial horizontal velocity,

ux=u−−√2−u2y=(125)2−(100)2−−−−−−−−−−−−√=75m/s

Tiem talen to reach the top of the hill,

t=2h−−√g=2×500−−−−−−√10)=10s

Horizontal distance covered in 10s=ux×t

=75×10=750m

So, canon has to be moved through a dustance

=800−650=50m

Time taken to move the canon through 50 m oh ground =502=25s

Total time taken by packet to reach ground the hill =25+10+10=45s