Question

A bomber plane moves horizontally with a speed of $500m/s$and a bomb released from it, strikes the ground in $10s$. Angle at which it strikes the ground will be $\left( {g = 10m/{s^2}} \right)$
A. ${\tan ^{ - 1}}\left( {\dfrac{1}{5}} \right)$
B. ${\tan ^{ - 1}}\left( {\dfrac{1}{2}} \right)$
C. ${\tan ^{ - 1}}\left( 1 \right)$
D. ${\tan ^{ - 1}}\left( 5 \right)$

Answer 1

Hint: First calculate the vertical component of velocity of the bomb using the equation of motion and then find the ratio of the vertical component to the horizontal component of velocity of the bomb. Then using suitable formula, obtain the angle.

Formula used
$v = u + at$ where $v$ is the final velocity, $u$ is the initial velocity, $t$ is the time taken and $a$ is the acceleration.
$\tan \theta = \dfrac{{{v_v}}}{{{v_h}}}$ where ${v_v}$ is the vertical component of velocity and ${v_h}$ is the horizontal component of velocity.

Complete step by step answer
This problem can be solved by implementing the equations of motions.
These equations of motions describe the behavior of a physical system based on their initial velocity, final velocity, acceleration and time.
Motion can be classified into two basic types- dynamics and kinematics.
In dynamics the forces and energies of the particles are taken into account. Whereas in kinematics only the position and time of the particle are taken into consideration.
The three main equations of motions are
$v = u + at$
${v^2} = {u^2} + 2aS$
$S = ut + \dfrac{1}{2}a{t^2}$
Where $v$ is the final velocity, $u$ is the initial velocity, $t$ is the time taken, $a$ is the acceleration and $S$ is the displacement of the body.
Initially the horizontal velocity of the bomb is $500m/s$
The vertical velocity of the bomb is calculated using the equation
$v = u + at$
Taking the initial vertical velocity to be $0$
So, $\begin{gathered}
  v = at \\
   \Rightarrow v = 10 \times 10m/s \\
\end{gathered} $
$ \Rightarrow v = 100m/s$
So the angle at which the bomb strikes the ground will be
$\tan \theta = \dfrac{{{v_v}}}{{{v_h}}}$ where ${v_v}$ is the vertical component of velocity and ${v_h}$ is the horizontal component of velocity.
$\begin{gathered}
  \tan \theta = \dfrac{{100}}{{500}} \\
   \Rightarrow \tan \theta = \dfrac{1}{5} \\
   \Rightarrow \theta = {\tan ^{ - 1}}\left( {\dfrac{1}{5}} \right) \\
\end{gathered} $

Therefore, the correct option is A.

Note: The equations of motion give us a comprehensive idea about the behavior of a body in motion. They are based on the three Newton’s laws of motion.