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Question: A ball is dropped from a building of height \({{45 m}}\). Simultaneously another ball is thrown up w...

A ball is dropped from a building of height 45m{{45 m}}. Simultaneously another ball is thrown up with a speed 40m/s{{40 m/s}}. The relative speed of the balls as a function of time is:
A) 20 m/s
B) 40 m/s
C) 30 m/s
D) 0 m/s

Explanation

Solution

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 40m/s{{40 m/s}}. Use the first equation of motion i.e. v=u+at{{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 vAB=vAvB{{{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 45m{{45 m}}
Speed of second ball is 40m/s{{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
uA=0{{{u}}_{{A}}}{{ = 0}}
Initial speed of ball B is given by
uB=40ms1{{{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{{v = u + at}}
For ball “A”
vA=uA+aAt\Rightarrow {{{v}}_{{A}}}{{ = }}{{{u}}_{{A}}}{{ + }}{{{a}}_{{A}}}{{t}}
On substituting the values we get
vA=0+gt\Rightarrow {{{v}}_{{A}}}{{ = 0 + g t}}
Where g = acceleration due to gravity
For ball “B”
vB=uB+aBt\Rightarrow {{{v}}_{{B}}}{{ = }}{{{u}}_{{B}}}{{ + }}{{{a}}_{{B}}}{{t}}
On substituting values, we get
vB=40+gt\Rightarrow {{{v}}_{{B}}}{{ = - 40 + g t}}
Relative speed of ball A with respect to ball B is given by,
vAB=vAvB\Rightarrow {{{v}}_{{{AB}}}}{{ = }}{{{v}}_{{A}}}{{ - }}{{{v}}_{_{{B}}}}
On substituting the values, we get
vAB=gt(40+gt) vAB=40ms1 \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 vAB=40ms1{{{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.