A hill is 500m high. Supplies are to be sent across the hill using...

Given, height of the hill (h) = 500 m

u = 125 m/s

To cross the hill, the vertical component of the velocity should be sufficient to cross such height.

∴ {\text{u}}_{\text{y}} ≥ \sqrt{2 \text{gh}}

≥ \sqrt{2 × 1 0 × 5 0 0}

≥ 1 0 0 \text{m/s}

But ${\text{u}}^{2} = {\text{u}}_{\text{x}}^{2} + {\text{u}}_{\text{y}}^{2}$

∴ Horizontal component of initial velocity

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

$= \sqrt{{\left(1 2 5\right)}^{2} - {\left(1 0 0\right)}^{2}}$

$= 7 5 \text{m} / \text{s}$

Time taken to reach the top of the hill

$\text{t} = \sqrt{\dfrac{2 \text{h}}{\text{g}}} = \sqrt{\dfrac{2 × 5 0 0}{1 0}} = 1 0 \text{s}$

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

Horizontal distance travelled in 10 s

$\text{x} = {\text{u}}_{\text{x}} × \text{t}$

$= 7 5 × 1 0$

$= 7 5 0 \text{m}$

∴ Distance through which canon has to be moved

= 800 - 750

= 50 m

Speed with which canon can move = 2 m/s

∴ Time taken by canon $= \dfrac{5 0}{2}$

{\text{t}}^{″} = 2 5 \text{s}

∴ Total time taken by a packet to reach on the ground

= {\text{t}}^{″} + \text{t} + {\text{t}}^{′}

$= 2 5 + 1 0 + 1 0$

$= 4 5 \text{s}$ .