A player throws a ball upwards with an i | Class 11 Physics Chapter Motion in a straight Line, Motion in a straight Line NCERT Solutions

Welcome to the NCERT Solutions for Class 11 Physics - Chapter Motion in a straight Line. This page offers a step-by-step solution to the specific question from Exercise 1, Question 10: . With detailed answers and explanations for each chapter, students can strengthen their understanding and prepare confidently for exams. Ideal for CBSE and other board students, this resource will simplify your study experience.

Question 10:

A player throws a ball upwards with an initial speed of 29.4 m s–1. What is the direction of acceleration during the upward motion of the ball? What are the velocity and acceleration of the ball at the highest point of its motion?

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. 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).

 

Answer:

Irrespective of the direction of the motion of the ball, acceleration (which is actually acceleration due to gravity) always acts in the downward direction towards the centre of the Earth.

At maximum height, velocity of the ball becomes zero. Acceleration due to gravity at a given place is constant and acts on the ball at all points (including the highest point) with a constant value i.e., 9.8 m/s2.

During upward motion, the sign of position is positive, sign of velocity is negative, and sign of acceleration is positive. During downward motion, the signs of position, velocity, and acceleration are all positive.

Initial velocity of the ball, u = 29.4 m/s

Final velocity of the ball, v = 0 (At maximum height, the velocity of the ball becomes zero)

Acceleration, a = – g = – 9.8 m/s2

From third equation of motion, height (s) can be calculated as:  

v2  -  u2  = 2gs

 s = v2  -  u2 / 2g

= (0)2 - (29.4)2 / 2 x (-9.8) = 44.1 m


From first equation of motion, time of ascent (t) is given as:  

v = u + at

t  =  v - u / a

= -29.4 / -9.8  = 3s


Time of ascent = Time of descent

Hence, the total time taken by the ball to return to the player’s hands = 3 + 3 = 6 s.


Study Tips for Answering NCERT Questions:

NCERT questions are designed to test your understanding of the concepts and theories discussed in the chapter. Here are some tips to help you answer NCERT questions effectively:

  • Read the question carefully and focus on the core concept being asked.
  • Reference examples and data from the chapter when answering questions about Motion in a straight Line.
  • Review previous year question papers to get an idea of how such questions may be framed in exams.
  • Practice answering questions within the time limit to improve your speed and accuracy.
  • Discuss your answers with your teachers or peers to get feedback and improve your understanding.

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Comments

  • Pratik Chavan
  • Aug 29, 2019

Nice


  • Diya
  • Jun 30, 2019

Why is velocity negative during upward motion?


Add Comment

Welcome to the NCERT Solutions for Class 11 Physics - Chapter . This page offers a step-by-step solution to the specific question from Excercise 1 , Question 10: A player throws a ball upwards with an initial speed of 29.4 m s–1. What is the direction of a....