**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 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.**

**Solution:**

Given,

Speed of packets = 125 m/s

Height of the hill = 500 m

Distance between the canon and the foot of the hill, d = 800 m

The vertical component of the velocity should be minimum so that the time taken to cross the hill will be the shortest.

uy = √2gh ≥ √(2)(10)(500) ≥ 100 m/s

But,

$u^{2}=u_{x}^{2}+u_{y}^{2}$

Therefore, horizontal component of the initial velocity,

$u_{x}=\sqrt{u^{2}-u_{y}^{2}}$

$\mathrm{u}_{\mathrm{x}}=75 \mathrm{~m} / \mathrm{s}$

Time taken by the packet to reach the top of the hill,

$t=\sqrt{\frac{2 h}{g}}$

$t=10 s$

Time taken to reach the ground from the top of the hill = t’ = t = 10s

Horizontal distance covered in 10s = (75)(10) = 750 m

Therefore, time taken by canon = 50/2 = 25 s

Therefore, total time taken by a packet to reach the ground = 25 + 10 + 10 = 45 s.