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 800m 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 m/s2.
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 800m 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 m/s2.
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1 Answer
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This is a Long Answer Type Question as classified in NCERT Exemplar
Explanation- speed of jackets = 125m/s
Height of hill = 500m
To cross the hill vertical component of velocity should be grater than this value uy=
So u2= ux2+uy2
Horizontal component of initial velocity ux =
Time taken to reach the top of hill t=
Time taken to reach the ground in 10 sec = 75 (10)= 750m
Distance through which the canon has to be moved =800-750=50m
Speed with which canon can move = 2m/s
Time taken canon = 50/2= 25s
Total time t= 25+10+10= 45s
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Please find the solution below:
after 10 kicks,
v? = 3tî v? = 24cos 60°î + 24sin 60°? = 12î + 12√3?
v? = v? – v? = (12 – 3t)î + 12√3?
It is minimum when 12 - 3t = 0 ⇒ t = 4sec
ω = θ² + 2θ
α = (ωdω)/dθ = (θ² + 2θ) (2θ + 2)
At θ = 1rad.
ω = 3rad/s and α = 12rad/s²
a? = αR = 12 m/s² a? = ω²R = 9 m/s² A? = √ (a? ² + a? ²) = 15 m/s²
a? = v? ²/4r
a_A? = (v? ²/r²) × r = v? ²/r
a_A = 3v? ²/4r
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