
An aeroplane is flying horizontally with a velocity of 216 km/hr and at a height of 1960 m. when it is vertically above a point A on the ground, a bomb is released from it. The bomb strikes the ground at point B. Find the distance AB.
Answer
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Hint: In these types of questions you should have prior knowledge about the distance and displacement and also remember that the velocity of the bomb will be the same as the velocity of aeroplane, use this information to approach towards the solution.
According to the given information velocity of aeroplane is 216 km/hr which is at height of 1960 m above from ground, since we know that when a bomb is released from the plane it will have the same velocity as the velocity of aeroplane since due to the bomb is thrown from height at speed of 216km/hr it will experience a gravitational force due to which the bomb will be in a projectile motion and AB will be the horizontal distance travelled by the bomb.
So velocity of bomb is equal to = 216 km/hr = 216 $ \times $$\dfrac{5}{{18}}$ = 60 m/s
Now we know that bomb is under free fall therefore it will take some time to reach at ground
Let t be the time taken by the bomb to reach at ground
So the time taken by bomb to reach ground will be $t = \sqrt {\dfrac{{2H}}{g}} $
Substituting the given values in the above equation we get
$t = \sqrt {\dfrac{{2 \times 1960}}{{10}}} $
$ \Rightarrow $$t = 14\sqrt 2 $
We know that since the bomb will cover the horizontal distance which can be calculated by the formula distance = speed $ \times $ time
Since AB is the distance travelled by the bomb
Substituting the values in the above equation we get
AB = 60 $ \times $ $14\sqrt 2 $
$ \Rightarrow $ AB = $840\sqrt 2 $ m
Therefore the distance AB is equal to $840\sqrt 2 $ m.
Note: In the above question we used the terms “distance and displacement” there is so much difference between the both terms which can be explained as distance represents the total measurement of path followed by any object whereas the displacement is the measurement of the smallest path followed by the object a displacement of a body can be negative whereas distance can’t be negative, distance is a vector quantity whereas the displacement is a vector quantity.
According to the given information velocity of aeroplane is 216 km/hr which is at height of 1960 m above from ground, since we know that when a bomb is released from the plane it will have the same velocity as the velocity of aeroplane since due to the bomb is thrown from height at speed of 216km/hr it will experience a gravitational force due to which the bomb will be in a projectile motion and AB will be the horizontal distance travelled by the bomb.
So velocity of bomb is equal to = 216 km/hr = 216 $ \times $$\dfrac{5}{{18}}$ = 60 m/s
Now we know that bomb is under free fall therefore it will take some time to reach at ground
Let t be the time taken by the bomb to reach at ground
So the time taken by bomb to reach ground will be $t = \sqrt {\dfrac{{2H}}{g}} $
Substituting the given values in the above equation we get
$t = \sqrt {\dfrac{{2 \times 1960}}{{10}}} $
$ \Rightarrow $$t = 14\sqrt 2 $
We know that since the bomb will cover the horizontal distance which can be calculated by the formula distance = speed $ \times $ time
Since AB is the distance travelled by the bomb
Substituting the values in the above equation we get
AB = 60 $ \times $ $14\sqrt 2 $
$ \Rightarrow $ AB = $840\sqrt 2 $ m
Therefore the distance AB is equal to $840\sqrt 2 $ m.
Note: In the above question we used the terms “distance and displacement” there is so much difference between the both terms which can be explained as distance represents the total measurement of path followed by any object whereas the displacement is the measurement of the smallest path followed by the object a displacement of a body can be negative whereas distance can’t be negative, distance is a vector quantity whereas the displacement is a vector quantity.
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