Answer
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Hint: First of all write the given quantities. Initial speed of the ball which is dropped from a building is zero. Initial velocity of the second ball which is thrown up is ${{40 m/s}}$. Use the first equation of motion i.e. ${{v = u + at}}$. Using the equation of motion, find out the final velocity of the first ball and then find out the final velocity of the second ball. For finding the relative speed of the balls, substitute the values of final velocities in formula ${{{v}}_{{{AB}}}}{{ = }}{{{v}}_{{A}}}{{ - }}{{{v}}_{_{{B}}}}$.
Complete step by step solution:
Given: There are two balls, on is dropped from a building and the other ball is thrown upwards.
The height of the building is ${{45 m}}$
Speed of second ball is ${{40 m/s}}$
To find: The relative speed of both the balls
Let ball “A” is dropped from building and ball “B” is thrown up
Initial speed of ball A is given by
${{{u}}_{{A}}}{{ = 0}}$
Initial speed of ball B is given by
${{{u}}_{{B}}}{{ = - 40 m}}{{{s}}^{{{ - 1}}}}$(as the direction of motion of the ball is opposite to the direction of acceleration due to gravity)
Using equation of motion we have
${{v = u + at}}$
For ball “A”
$\Rightarrow {{{v}}_{{A}}}{{ = }}{{{u}}_{{A}}}{{ + }}{{{a}}_{{A}}}{{t}}$
On substituting the values we get
$\Rightarrow {{{v}}_{{A}}}{{ = 0 + g t}}$
Where g = acceleration due to gravity
For ball “B”
$\Rightarrow {{{v}}_{{B}}}{{ = }}{{{u}}_{{B}}}{{ + }}{{{a}}_{{B}}}{{t}}$
On substituting values, we get
$\Rightarrow {{{v}}_{{B}}}{{ = - 40 + g t}}$
Relative speed of ball A with respect to ball B is given by,
$\Rightarrow {{{v}}_{{{AB}}}}{{ = }}{{{v}}_{{A}}}{{ - }}{{{v}}_{_{{B}}}}$
On substituting the values, we get
$
\Rightarrow {{{v}}_{{{AB}}}}{{ = gt - ( - 40 + gt)}} \\
\therefore {{{v}}_{{{AB}}}}{{ = 40 m}}{{{s}}^{ - 1}}$
Thus, the relative speed of the balls as a function of time is ${{{v}}_{{{AB}}}}{{ = 40 m}}{{{s}}^{ - 1}}$.
Therefore, option (B) is the correct choice.
Note: When the ball is thrown upwards then the value of velocity becomes negative because of the fact that the acceleration due to gravity acts downwards (i.e. towards the surface of the earth). When the ball is thrown downwards then the value of velocity becomes positive because of the fact that the acceleration due to gravity acts downwards and which is in the direction of the ball.
Complete step by step solution:
Given: There are two balls, on is dropped from a building and the other ball is thrown upwards.
The height of the building is ${{45 m}}$
Speed of second ball is ${{40 m/s}}$
To find: The relative speed of both the balls
Let ball “A” is dropped from building and ball “B” is thrown up
Initial speed of ball A is given by
${{{u}}_{{A}}}{{ = 0}}$
Initial speed of ball B is given by
${{{u}}_{{B}}}{{ = - 40 m}}{{{s}}^{{{ - 1}}}}$(as the direction of motion of the ball is opposite to the direction of acceleration due to gravity)
Using equation of motion we have
${{v = u + at}}$
For ball “A”
$\Rightarrow {{{v}}_{{A}}}{{ = }}{{{u}}_{{A}}}{{ + }}{{{a}}_{{A}}}{{t}}$
On substituting the values we get
$\Rightarrow {{{v}}_{{A}}}{{ = 0 + g t}}$
Where g = acceleration due to gravity
For ball “B”
$\Rightarrow {{{v}}_{{B}}}{{ = }}{{{u}}_{{B}}}{{ + }}{{{a}}_{{B}}}{{t}}$
On substituting values, we get
$\Rightarrow {{{v}}_{{B}}}{{ = - 40 + g t}}$
Relative speed of ball A with respect to ball B is given by,
$\Rightarrow {{{v}}_{{{AB}}}}{{ = }}{{{v}}_{{A}}}{{ - }}{{{v}}_{_{{B}}}}$
On substituting the values, we get
$
\Rightarrow {{{v}}_{{{AB}}}}{{ = gt - ( - 40 + gt)}} \\
\therefore {{{v}}_{{{AB}}}}{{ = 40 m}}{{{s}}^{ - 1}}$
Thus, the relative speed of the balls as a function of time is ${{{v}}_{{{AB}}}}{{ = 40 m}}{{{s}}^{ - 1}}$.
Therefore, option (B) is the correct choice.
Note: When the ball is thrown upwards then the value of velocity becomes negative because of the fact that the acceleration due to gravity acts downwards (i.e. towards the surface of the earth). When the ball is thrown downwards then the value of velocity becomes positive because of the fact that the acceleration due to gravity acts downwards and which is in the direction of the ball.
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