3.12 A ball is dropped from a height of 90 m on a floor. At each collision with the floor, the ball loses one tenth of its speed. Plot the speed-time graph of its motion between t = 0 to 12 s.

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6 months ago

Given data: 

 Initial height, h = 90 m 

 Acceleration due to gravity, g = 9.8 ms2 

 Speed lost at each collision = 1/10 of its speed 

 time interval to be considered: 0 to 12 secs. 

 Step 1: Time to fall from 90 m  

h = ½  gt2 ⇒ 90 = 21 * 9.8 * t2 ⇒ t2 = 9.8180 ≈ 18.37 ⇒ t ≈ 4.29 seconds 

Step 2: Speed bef

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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 solutions class 11th 2023

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