Question

A reckless drunk is playing with a gun in an airplane that is going to east at \[500km/h\]. The drunk shoots the gun straight up at the ceiling of the plane. The bullet leaves the gun at a speed of \[100km/h\]. According to an observer on earth, what angle does the bullet make with the vertical?

Answer 1

Hint:In this question, we have two components of velocity of different objects which are at right angles to each other. So, firstly we find the resultant velocity of two components of different velocity, this resultant velocity will make two angles with the components of velocities. One angle is with horizontal components and second with vertical components.

Formula used:When two components of a physical quantity are given then resultant magnitude of that physical quantity is
Resultant$ = \sqrt {{{\left( {horizontal} \right)}^2} + {{\left( {vertical} \right)}^2}} $
If velocity is physical quantity then-
Resultant velocity$ = \sqrt {{{\left( {{v_h}} \right)}^2} + {{\left( {{v_v}} \right)}^2}} $
Where ${v_h}$ and ${v_v}$ are the horizontal and vertical components of velocity respectively.
The angle that the resultant velocity makes with the horizontal component is-
$\tan \theta = \dfrac{{perpendicular}}{{base}}$

Complete step by step solution: -
According to the question, we know that a drunk fired a bullet from a gun in upward direction in a moving airplane with \[500km/h\] in the east direction. If the velocity of the fixed bullet is \[100km/h\] having a direction at right angle with the moving airplane. So, these velocities are called horizontal and vertical components. Now, the resultant velocity of the bullet is seen by an observer is-
$v = \sqrt {{{\left( {{v_h}} \right)}^2} + {{\left( {{v_v}} \right)}^2}} $
Putting the values of ${v_h}$ and ${v_v}$, we get-
$
   \Rightarrow v = \sqrt {{{\left( {500} \right)}^2} + {{\left( {1000} \right)}^2}} \\
   \Rightarrow v = \sqrt {250000 + 1000000} \\
   \Rightarrow v = \sqrt {1250000} \\
   \Rightarrow v = 500\sqrt 5 km/h \\
$
Now, this resultant velocity will make two angles – one angle is associated with the horizontal component and the second angle is associated with the vertical component. Observer will see angle second angle which is the angle associated with vertical component:
$
  \tan \phi = \dfrac{{horizontal}}{{vertical}} \\
   \Rightarrow \tan \phi = \dfrac{{500}}{{1000}} \\
   \Rightarrow \tan \phi = \dfrac{1}{2} \\
   \Rightarrow \phi = {\tan ^{ - 1}}\left( {\dfrac{1}{2}} \right) \\
   \Rightarrow \phi = {26.56^ \circ } \\
$
The value of ${\tan ^{ - 1}}\left( {\dfrac{1}{2}} \right)$ is ${26.56^ \circ }$. Hence, the observer will see the angle of ${26.56^ \circ }$ which the bullet makes with the vertical component.

Note: -When the angle, which is resultant velocity, is horizontally asked in the question, then there we used the same formula for another angle. If an observer sees this resultant velocity at rest so it will be at a vertical resultant angle.
It is only seen by an observer meaning the bullet does not actually make ${26.56^ \circ }$ of angle because the drunk fired bullet directed straight up to the ceiling. Which become at right angles with velocity of the airplane.