# Class 11 Physics Motion in a straight Line: NCERT Solutions for Question 10

This page focuses on the detailed Motion in a straight Line question answers for Class 11 Physics Motion in a straight Line, addressing the question: '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).  '. The solution provides a thorough breakdown of the question, highlighting key concepts and approaches to arrive at the correct answer. This easy-to-understand explanation will help students develop better problem-solving skills, reinforcing their understanding of the chapter and aiding in exam preparation.
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).

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.