Question

At the height of $80m$, an aeroplane is moving with $150m{s}^{-1}$. A bomb is dropped from it, so as to hit a target. At what distance from the target should bomb be dropped?

($g=10m{s}^{-2}$).

Answer 1

Step 1: Given

Height of aeroplane, $h=80m$

Speed of aeroplane, $v_H=150m{s}^{-1}$

Acceleration due to gravity, $g=10m{s}^{-2}$

Step 2: Formula used and calculation

Step 2: Calculation

Time taken to reach the ground or target is given by

$$\begin{gathered} h=ut+\frac{1}{2}a{t}^2 \\ \Rightarrow h=0\times t+\frac{1}{2}\times g\times t^2 \\ \Rightarrow t=\sqrt{\frac{2h}{g}} \\ \Rightarrow t=\sqrt{\frac{2\times 80m}{10m{s}^{-2}}} \\ \Rightarrow t=4s \end{gathered}$$

Horizontal distance, $BC$, covered by the bomb is given by

$$\begin{gathered} BC=v_H\times t \\ \Rightarrow BC=150m{s}^{-1}\times 4s \\ \Rightarrow BC=600m \end{gathered}$$

Now,

$$\begin{gathered} AC=\sqrt{B{C}^2+A{B}^2} \\ \Rightarrow AC=\sqrt{{600}^2+{80}^2} \\ \Rightarrow AC=605.3m \end{gathered}$$

Hence, the horizontal distance between aeroplane and dropping point of plane is $600m$.

Hence, actual distance between aeroplane and dropping point of plane is $605.3m$.