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Home » 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 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 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 m/s2.
CBSE | Class 11 | Motion in a Plane | NCERT | Physics
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 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.

Answer:

The speed of packets is 125 m/s, the height of the hill is 500 m, and the distance between the cannon and the foot of the hill is 800 m, according to the problem.

Ideally, the vertical component of the velocity should be kept to a minimum so that it spans the height of the slope in the least amount of time.

uy=2gh≥2×10×500≥100m/s
u_{y}=\sqrt{2 g h} \geq \sqrt{2 \times 10 \times 500} \geq 100 m / s

But u^{2}=u_{x}^{2}+u_{y}^{2}
So, Horizontal component of initial velocity,

ux=u2–uy2=(125)2–(100)2=75m/su_{x}=\sqrt{u^{2}-u_{y}^{2}}=\sqrt{(125)^{2}-(100)^{2}}=75 m / s

The amount of time it took the packet to get to the top of the slope,

t=2hg=2×50010=10st=\sqrt{\frac{2 h}{g}}=\sqrt{\frac{2 \times 500}{10}}=10 s

Time taken to reach the ground from the top of the hill t^{\prime}=t=10 s. Horizontal distance travelled in 10 s,

x=ux×t=75×10=750mx=u_{x} \times t=75 \times 10=750 m

The distance through which the cannon must be moved is equal to 800 – 750 = 50 metres. The maximum speed at which a canon may move is 2 metres per second.

\therefore Time taken by canon \frac{50}{2} \Rightarrow t^{\prime \prime}=25 s
\therefore Total time taken by a packet to reach on the ground =t^{\prime \prime}+t+t^{\prime}

=25+10+10=45s=25+10+10=45 s
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