Question

A bomb is dropped from an aeroplane when it is directly above a target at a height of 1000 m. The aeroplane is moving horizontally at a speed of 500 km/h. By how much distance will the bomb miss the target?

Answer 1

Step 1: Given data

1. The velocity of the aeroplane is $v=500km/h$.
2. The distance between the target and the aeroplane is $d=1000m.$

Concepts and formulae

1. The plane is moving horizontally, suppose it is along the x-axis. So the initial velocity along the y axis (vertically) is zero, i.e, $u_y=0.$
2. We know the formulae of kinematics, $s=ut+\frac{1}{2}a{t}^2$, where s and a are the distance traveled by a body at time t and acceleration a respectively.

Diagram

Step 2: Find the distance

Let, x be the distance between the target site and the actual site where the bomb lands..

We know, that the initial velocity along the vertical direction is zero, i.e, $u_y=0.$. So,

$$\begin{gathered} S_y=u_{y}t+\frac{1}{2}g{t}^2 \\ or1000=0+\frac{1}{2}\times (9.8)\times t^2 \\ or{t}^2=\frac{1000\times 2}{9.8}=204.08 \\ ort=14.29sec. \end{gathered}$$

So, the time of flying of the bomb is $t=14.29sec.$

Step 3: Find the velocity of the airplane

$$\begin{gathered} v=\frac{s}{t} \\ v=\frac{500km}{1h}=\frac{500000m}{3600sec}=138.89m/s \\ orv=138.89m/s. \end{gathered}$$

Again,

$$\begin{gathered} S=vt \\ orx=138.9\times 14.29 \\ orx=1984.8meter. \end{gathered}$$

Therefore, the target will falls 1984.8 m away from the target.