
At a 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 the bomb be dropped? $g = 10m{s^{ - 2}}$.
Answer
532.2k+ views
Hint: 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,$s = ut + \dfrac{1}{2}a{t^2}$ 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 $150m{s^{ - 1}}$ and the acceleration due gravity is$g = 10m{s^{ - 2}}$.
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
$ \Rightarrow s = ut + \dfrac{1}{2}a{t^2}$
Here replace$s = 180m$, $a = g = 10m{s^{ - 2}}$ also the initial velocity$u = 0$.
$ \Rightarrow 180 = \dfrac{1}{2}\left( {10} \right){t^2}$
$ \Rightarrow 360 = 10{t^2}$
$ \Rightarrow 36 = {t^2}$
$ \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 $150m{s^{ - 1}}$ and the time taken by the bomb to hit the target is 6sec.
The formula of speed is given by,
$ \Rightarrow {\text{speed}} = \dfrac{{{\text{distance}}}}{{{\text{time}}}}$
$ \Rightarrow 150 = \dfrac{{{\text{distance}}}}{6}$
$ \Rightarrow {\text{distance}} = 150 \times 6$
$ \Rightarrow {\text{distance}} = 900m$.
The distance at which the bomb should be dropped is $d = 900m$.
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.
Formula used: The formula of the Newton’s law of motion are,$s = ut + \dfrac{1}{2}a{t^2}$ 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 $150m{s^{ - 1}}$ and the acceleration due gravity is$g = 10m{s^{ - 2}}$.
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
$ \Rightarrow s = ut + \dfrac{1}{2}a{t^2}$
Here replace$s = 180m$, $a = g = 10m{s^{ - 2}}$ also the initial velocity$u = 0$.
$ \Rightarrow 180 = \dfrac{1}{2}\left( {10} \right){t^2}$
$ \Rightarrow 360 = 10{t^2}$
$ \Rightarrow 36 = {t^2}$
$ \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 $150m{s^{ - 1}}$ and the time taken by the bomb to hit the target is 6sec.
The formula of speed is given by,
$ \Rightarrow {\text{speed}} = \dfrac{{{\text{distance}}}}{{{\text{time}}}}$
$ \Rightarrow 150 = \dfrac{{{\text{distance}}}}{6}$
$ \Rightarrow {\text{distance}} = 150 \times 6$
$ \Rightarrow {\text{distance}} = 900m$.
The distance at which the bomb should be dropped is $d = 900m$.
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.
Recently Updated Pages
Why are manures considered better than fertilizers class 11 biology CBSE

Find the coordinates of the midpoint of the line segment class 11 maths CBSE

Distinguish between static friction limiting friction class 11 physics CBSE

The Chairman of the constituent Assembly was A Jawaharlal class 11 social science CBSE

The first National Commission on Labour NCL submitted class 11 social science CBSE

Number of all subshell of n + l 7 is A 4 B 5 C 6 D class 11 chemistry CBSE

Trending doubts
What is meant by exothermic and endothermic reactions class 11 chemistry CBSE

10 examples of friction in our daily life

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

What are Quantum numbers Explain the quantum number class 11 chemistry CBSE

