A ball is dropped from a building of height 45 m. Simultaneously another ball is thrown up with a speed 40 m/s. Calculate the relative speed of the balls as a function of time.

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7 months ago

This is a short answer type question as classified in NCERT Exemplar

When the ball dropped from the building u1=0, u2=40m/s

Velocity of the dropped ball after time t

V1=u1+gt

V1 = gt

For ball thrown up u2=40m/s

Velocity of the ball after time t

V2=u2-gt

=40-gt

Relative velocity =v1-v2

=gt- [-40-gt]=40m/s

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According to question, we can write

 10 = 1 2 a t 2 . . . . . . . . . . . . . . . ( 1 ) a n d                        

1 0 + x = 1 2 a ( 2 t ) 2 . . . . . . . . . . . . . . . . ( 2 )  

X = 1 2 A [ ( 2 t ) 2 t 2 ] = 3 ( 1 2 a t 2 ) = 3 0 m  

 

Average speed = 4 v 2 3 v

= 4 v 3

(d) Initial velocity  = - v j ˆ

Final velocity = v i ˆ

Change in velocity  = v i ˆ - ( - v j ˆ )

= v ( i ˆ + j ˆ ) Momentum gain is along i ˆ + j ˆ

 Force experienced is along i ˆ + j ˆ

  Force experienced is in North-East direction.

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Physics NCERT Exemplar Solutions Class 11th Chapter Three 2025

Physics NCERT Exemplar Solutions Class 11th Chapter Three 2025

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