Question

A bomber plane moves horizontally with speed of 600 m/s and a bomb is released from it, strikes the ground in 10 s. The angle with horizontal at which it strikes the ground will be
A. ${\tan ^{ - 1}}\left( {1/2} \right)$
B. ${\tan ^{ - 1}}\left( {1/6} \right)$
C. ${\tan ^{ - 1}}\left( {4/5} \right)$
D. ${\tan ^{ - 1}}\left( {3/4} \right)$

Answer 1

We see that the horizontal velocity of the bomb is the same as the horizontal velocity of the bomber plane as there is no force in the horizontal direction. In the vertical direction, due to acceleration due to gravity, the vertical component of the final velocity can be found using the first kinematic equation in vertical direction. You can make the simple vector construction of horizontal and vertical components of final velocity to determine the angle.

Formula used:
$v = u + at$
Here, v is the final velocity, u is the initial velocity, $a$ is the acceleration and t is the time.

Complete step by step answer:
We assume the horizontal and vertical velocity of the bomb when it strikes the ground is ${v_x}$ and ${v_y}$ respectively.
We can see there is no force in the horizontal direction to accelerate the bomb. Therefore, the horizontal component of the ball will be the same as the horizontal velocity of the bomber plane.
We have given that the bomb is released from the plane that means the initial velocity of the bomb is zero. We can use the first kinematic equation to determine the vertical component of the velocity of the bomb.
${v_y} = {u_y} + gt$
Here, ${u_y}$ is the vertical component of initial velocity, g is the acceleration due to gravity and t is the time.
Since the vertical component of initial velocity is zero, we can write the above equation as,
${v_y} = gt$
Now, we can substitute $10\,m/{s^2}$ for g and 10 s for t in the above equation.
${v_y} = \left( {10\,m/{s^2}} \right)\left( {10\,s} \right)$
$\Rightarrow {v_y} = 100\,m/s$
So, we now have horizontal and vertical components of the final velocity of the bomb. This can be represented by the vector diagram as follows,

From the geometry of the above figure, we can express the angle made by the bomb with horizontal $\theta$ as below,
$\tan \theta = \dfrac{{{v_y}}}{{{v_x}}}$
$\Rightarrow \theta = {\tan ^{ - 1}}\left( {\dfrac{{{v_y}}}{{{v_x}}}} \right)$
Now, we have to substitute ${v_y} = 100\,m/s$ and ${v_x} = 600\,m/s$ in the above equation.
$\theta = {\tan ^{ - 1}}\left( {\dfrac{{100}}{{600}}} \right)$
$\therefore \theta = {\tan ^{ - 1}}\left( {\dfrac{1}{6}} \right)$
This is the angle made by the bomb with horizontal when it strikes the ground.

So, the correct answer is “Option B”.

Note:
While using the first kinematic equation, $v = u + at$ for a body moving towards the ground, both final velocity and acceleration due to gravity are in the same direction. Therefore, you can substitute $+ g$ for acceleration a in the above equation. For the body moving upward, the final velocity and acceleration due to gravity are in the opposite direction. Therefore, in such cases you should substitute $- g$ for acceleration.