Question

A cannon $B$ and a target are $0.98\text{km}$ apart located at the same level. Find the time taken to hit the target, if the speed of the shell fired by the cannon is $98\text{m/s}$. Neglect the air drag and take $g=9.8{\text{m/s}}^2$.

• A. $10\sqrt{3}\text{s}$
• B. $10\text{s}$
• C. $10\sqrt{2}\text{s}$
• D. $20\sqrt{2}\text{s}$

Answer 1

The correct option is C $10\sqrt{2}\text{s}$

In order to hit the target, the horizontal distance between the cannon and target must be equal to the range of the shell fired.
$R=0.98\text{km}=980\text{m}$
Given, initial speed of projection, $u=98\text{m/s}$
On using the formula
$R=\frac{u^2\sin 2\theta }{g}$
$\Rightarrow \sin 2\theta =\frac{R\times g}{u^2}=\frac{980\times 9.8}{98\times 98}$
$\Rightarrow \sin 2\theta =1$
Or, $2\theta ={90}^{\circ }$
$\therefore \theta ={45}^{\circ }$
Now, Time of flight is,
$T=\frac{2u\sin \theta }{g}=\frac{2\times 98\times \sin {45}^{\circ }}{9.8}$
$\therefore T=20\times \frac{1}{\sqrt{2}}=10\sqrt{2}\text{s}$