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Question: At a height of 80m, an aeroplane is moving with\(150m{s^{ - 1}}\). A bomb is dropped from it, so as ...

At a height of 80m, an aeroplane is moving with150ms−1150m{s^{ - 1}}. A bomb is dropped from it, so as to hit a target. At what distance from the target should the bomb be dropped? g=10ms−2g = 10m{s^{ - 2}}.

Explanation

Solution

As the bomb goes and hits the target it should be understood that there are two motions one is vertical and one is horizontal and both of them should take the same time from dropping off the bomb to hitting the target. Newton's law of motion relations can be used here to find the correct answer to this problem.
Formula used: The formula of the Newton’s law of motion are,s=ut+12at2s = ut + \dfrac{1}{2}a{t^2} where distance is s, the initial velocity is u, the acceleration is a and the time taken is t.

Complete Step by step solution:
As it is given that the vertical height of the plane is 80m also it is given that the speed of the plane is 150ms−1150m{s^{ - 1}} and the acceleration due gravity isg=10ms−2g = 10m{s^{ - 2}}.
Let us calculate the time that the bomb takes to hit the target.
For motion in vertical direction.
Apply Newton’s law of motion’s relation
⇒s=ut+12at2\Rightarrow s = ut + \dfrac{1}{2}a{t^2}
Here replaces=180ms = 180m, a=g=10ms−2a = g = 10m{s^{ - 2}} also the initial velocityu=0u = 0.
⇒180=12(10)t2\Rightarrow 180 = \dfrac{1}{2}\left( {10} \right){t^2}
⇒360=10t2\Rightarrow 360 = 10{t^2}
⇒36=t2\Rightarrow 36 = {t^2}
⇒t=6s\Rightarrow t = 6s
This is the time taken to hit the target is 6sec.
For horizontal motion,
The speed of the aeroplane is equal to 150ms−1150m{s^{ - 1}} and the time taken by the bomb to hit the target is 6sec.
The formula of speed is given by,
⇒speed=distancetime\Rightarrow {\text{speed}} = \dfrac{{{\text{distance}}}}{{{\text{time}}}}
⇒150=distance6\Rightarrow 150 = \dfrac{{{\text{distance}}}}{6}
⇒distance=150×6\Rightarrow {\text{distance}} = 150 \times 6
⇒distance=900m\Rightarrow {\text{distance}} = 900m.

The distance at which the bomb should be dropped is d=900md = 900m.

Note: It is advisable for students to remember all the three relations and understand all the relations in Newton's law of motion as they are extremely important in solving these types of problems. The sign of acceleration of the horizontal motion of the bomb is negative as there is deceleration in the motion of the bomb.