Question

A ball is projected horizontally with speed \[10\,{\text{m/s}}\] from the top of the building of height \[20\,{\text{m}}\]. The angle made by the velocity vector with the horizontal when ball will hit the ground will be (Take \[g = 10\,{\text{m/}}{{\text{s}}^2}\])
A. \[{\tan ^{ - 1}}\left( 2 \right)\]
B. \[{\tan ^{ - 1}}\left( {\dfrac{1}{2}} \right)\]
C. \[{\tan ^{ - 1}}\left( 4 \right)\]
D. \[{\tan ^{ - 1}}\left( {\dfrac{2}{3}} \right)\]

Answer 1

Hint:Use the kinematic equation for the final velocity of an object. This equation gives the relation between the final velocity of the object, initial velocity of the object, acceleration of the object and displacement of the object. Using this kinematic equation, calculate the vertical component of final velocity of the ball and use the initial horizontal velocity as horizontal component of final velocity of the ball and calculate the required angle.

Formula used:
The kinematic equation for final velocity of an object in terms of its displacement \[s\] is
\[{v^2} = {u^2} + 2as\] …… (1)
Here, \[u\] is initial velocity of the object, \[v\] is final velocity of the object and \[a\] is acceleration of the object.

Complete step by step answer:
We have given that the initial horizontal velocity of the ball is \[10\,{\text{m/s}}\].
\[{u_x} = 10\,{\text{m/s}}\]
The height of the building from which the ball is projected is \[20\,{\text{m}}\].
\[h = 20\,{\text{m}}\]
We are asked to calculate the angle made by the velocity vector when the ball touches the ground.Let us first calculate the vertical component of the speed of the ball when it touches the ground.The initial vertical velocity of the ball is zero as it is projected horizontally.
\[{u_y} = 0\,{\text{m/s}}\]
Rewrite equation (1) for the vertical displacement of the ball.
\[v_y^2 = u_y^2 + 2gh\]
Substitute \[0\,{\text{m/s}}\] for \[{u_y}\], \[10\,{\text{m/}}{{\text{s}}^2}\] for \[g\] and \[20\,{\text{m}}\] for \[h\] in the above equation.
\[v_y^2 = {\left( {0\,{\text{m/s}}} \right)^2} + 2\left( {10\,{\text{m/}}{{\text{s}}^2}} \right)\left( {20\,{\text{m}}} \right)\]
\[ \Rightarrow v_y^2 = 400\]
\[ \Rightarrow {v_y} = 20\,{\text{m/s}}\]
Hence, the vertical component of velocity of the ball when it touches the ground is \[20\,{\text{m/s}}\].

Since the ball is projected horizontally from the top of the building, the initial horizontal velocity of the ball remains the same throughout the motion of the ball. As the ball is in free fall, only the vertical component of velocity of the ball changes and the horizontal component remains the same. Hence, the horizontal component of velocity of the ball when it touches the ground is \[10\,{\text{m/s}}\].
\[{v_x} = 10\,{\text{m/s}}\]
The angle made by the velocity vector of the ball with the horizontal is
\[\theta = {\tan ^{ - 1}}\left( {\dfrac{{{v_y}}}{{{v_x}}}} \right)\]
Substitute \[20\,{\text{m/s}}\] for \[{v_y}\] and \[10\,{\text{m/s}}\] for \[{v_x}\] in the above equation.
\[\theta = {\tan ^{ - 1}}\left( {\dfrac{{20\,{\text{m/s}}}}{{10\,{\text{m/s}}}}} \right)\]
\[ \therefore \theta = {\tan ^{ - 1}}\left( 2 \right)\]
Therefore, the angle made by the velocity vector when the ball touches the ground is \[{\tan ^{ - 1}}\left( 2 \right)\].

Hence, the correct option is A.

Note: The students should not forget to use the same initial horizontal velocity of the ball as the horizontal component of final velocity of the ball. This is because for an object in free fall the vertical component of velocity changes but the horizontal component of velocity remains the same throughout its motion.