3.10 A player throws a ball upwards with an initial speed of 29.4 m s–1.

(a) What is the direction of acceleration during the upward motion of the ball?

 

(b) What are the velocity and acceleration of the ball at the highest point of its motion?

 

(c) Choose the x = 0 m and t = 0 s to be the location and time of the ball at its highest point, vertically downward direction to be the positive direction of x axis, and give the signs of position, velocity and acceleration of the ball during its upward, and downward motion

 

(d) To what height does the ball rise and after how long does the ball return to the player's hands? (Take g = 9.8 m s–2 and neglect air resistance)

469 Views|Posted 8 months ago
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6 months ago

3.10

(a) The direction of the acceleration is downward during the upward motion of the ball because the acceleration is due to gravity. Gravity always pulls the object toward the centre of the earth. 

(b) Velocity: At the peak the velocity is zero because the ball momentarily stops before falling do

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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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