Physics - Kinematics Question with Solution | TestHub

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.

$\therefore$${\text{u}}_{\text{y}}\ge \sqrt{2\text{gh}}$

$\ge \sqrt{2\times 10\times 500}$

$\ge 10\mathrm{0 }\text{m/s}$

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

$\therefore$Horizontal component of initial velocity

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

$=\sqrt{{\left(125\right)}^2-{\left(100\right)}^2}$

$=75\text{ m}/\text{s}$

Time taken to reach the top of the hill

$\text{t}=\sqrt{\frac{2\text{h}}{\text{g}}}=\sqrt{\frac{2\times 500}{10}}=1\mathrm{0 }\text{s}$

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

Horizontal distance travelled in 10 s

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

$=75\times 10$

$=750\text{ m}$

$\therefore$Distance through which canon has to be moved

= 800 - 750

= 50 m

Speed with which canon can move = 2 m/s

$\therefore$Time taken by canon$=\frac{50}{2}$

${\text{t}}^{″}=2\mathrm{5 }\text{s}$

$\therefore$Total time taken by a packet to reach on the ground

$={\text{t}}^{″}+\text{t}+{\text{t}}^{\prime }$

$=25+10+10$

$=45\text{ s}$.