Question

An aeroplane flying horizontally at an altitude of 490 m with a speed of 180kmph drops a bomb. The horizontal distance at which it hits the ground is
(A) 500 m
(B) 1000 m
(C) 250 m
(D) 50 m

Answer 1

Hint
The bomb when dropped will also possess a horizontal speed of 180kmph, and will maintain that velocity until it lands. The time taken for the bomb to vertically fall to the ground is also the time taken for the bomb to travel the horizontal distance. So from here we can calculate the horizontal distance.
Formula used: In this solution we will be using the following formula,
 $ \Rightarrow x = {v_x}t $ where $ x $ is the horizontal distance travelled by an object, $ {v_x} $ is the horizontal velocity of the object and $ t $ is the time taken.
 $ \Rightarrow y = {v_{0x}}t + \dfrac{1}{2}g{t^2} $ where $ y $ is the vertical distance travelled, $ {v_{0y}} $ is the initial vertical velocity of the object, and $ g $ is the acceleration due to gravity.

Complete step by step answer
When the bomb was dropped, the aeroplane was flying at speed 180 kmph, hence the bomb must possess the same horizontal speed. The vertical speed however at the time of drop is zero, since the bomb was not fired vertically.
For horizontal distance travelled at time $ t $ of journey is given by
 $ x = {v_x}t $ where $ {v_x} $ is the horizontal velocity of the object.
 $ \Rightarrow {v_x} = 180kmph $ (since it must be equal to the speed of the aeroplane). To calculate we must convert to $ m/s $
 $ \Rightarrow {v_x} = 180 \times \dfrac{{1000m}}{{60 \times 60s}} = 50m/s $
Then by inserting value into $ x = {v_x}t $ we have
 $ \Rightarrow x = 50t $
Also, the vertical distance covered during this journey is given by
 $ \Rightarrow y = {v_{0x}}t + \dfrac{1}{2}g{t^2} $ where $ {v_{0y}} $ is the initial vertical speed of the object, and $ g $ is the acceleration due to gravity.
As mentioned above, initial vertical speed is zero, then the above equation reduces to
 $ \Rightarrow y = \dfrac{1}{2}g{t^2} $
For calculating the time taken to fall to the ground we have,
 $ \Rightarrow {t^2} = \dfrac{{2y}}{g} $
 $ \Rightarrow t = \sqrt {\dfrac{{2y}}{g}} $
Inserting the known values into the formula, we have
 $ \Rightarrow t = \sqrt {\dfrac{{2\left( {490} \right)}}{{9.8}}} = 10s $
Substituting the value of $ t $ into $ x = 50t $, we have
 $ \Rightarrow x = 50 \times 10 = 500m $
 $ \therefore x = 500m $.
Hence, the correct option is (A).

Note
In real life, the horizontal distance covered is less than estimated above. This is due to the air resistance that would act to oppose the bomb speed, thus reducing the speed of the bomb in time. Air resistance, like friction, always acts opposite to the direction of motion.