A bomb is dropped by an aero plane flying horizontally with a velocity $200 \mathrm{~km} / \mathrm{hr}$ and…
A bomb is dropped by an aero plane flying horizontally with a velocity $200 \mathrm{~km} / \mathrm{hr}$ and at a height of $980 \mathrm{~m}$. At the time of dropping a bomb, the distance of the aero plane from the target on the ground to hit directly is $\left(\mathrm{g}=9.8 \mathrm{~m} / \mathrm{s}^2\right.$ )
$\frac{\sqrt{2} \times 10^4}{9} \mathrm{~m}$
$\frac{10^4}{9} \mathrm{~m}$
$\frac{10^4}{9 \sqrt{2}} \mathrm{~m}$
$\frac{10^4}{18} \mathrm{~m}$
Solution
The plane is flying horizontally. Hence initial vertical component of the velocity is zero.
If it reaches the ground in time $t$, then
$\begin{aligned}
& \mathrm{h}=\frac{1}{2} \mathrm{gt}^2 \\
& \mathrm{t}^2=\frac{2 \mathrm{~h}}{\mathrm{~g}}=\frac{2 \times 980}{9.8}=200 \\
& \mathrm{t}=10 \sqrt{2} \mathrm{~s}
\end{aligned}$
The horizontal component of the velocity is
$\mathrm{V}=200 \mathrm{~km} / \mathrm{hr}=200 \times \frac{5}{18}=\frac{1000}{18} \mathrm{~m} / \mathrm{s}$
The horizontal distance to be covered is
$\mathrm{d}=\mathrm{Vt}=\frac{1000}{18} \times 10 \sqrt{2}=\frac{10^4}{9 \sqrt{2}} \mathrm{~m}$