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).
Given, the initial speed of the ball is 29.4 m / s .
(a)
When the ball is thrown vertically upwards, the only external force acting on the ball is gravitational force. Since the direction of action of gravitational force is in downward direction; therefore, the acceleration of the motion of the ball will be in the downward direction.
(b)
When the ball reaches its highest point the velocity of the ball at that instant is zero, and the acceleration of the ball is equal to the acceleration due to gravity, that is g = 9.8 m / s 2 .
(c)
Let the position of the ball at the highest point be x = 0 , time be t = 0 . Consider the downward motion of the ball as positive direction of x axis.
When the ball is in upward motion the sign of position and the acceleration of the ball is positive and the sign of the velocity of the ball is negative.
When the ball is in downward motion, the sign of position, velocity and the acceleration of the ball are positive.
(d)
Let u be the initial velocity of the ball, v be the final velocity, g be the acceleration due to gravity and s be the distance covered by the ball.
The third equation of motion is,
v 2 = u 2 − 2 g s
Substitute the required values in the above equation.
0 = ( − 29.4 m / s ) 2 − 2 ( 9.8 m / s 2 ) × s s = 44.1 m
Thus, the ball rises to a height of 44.1 m.
The first equation of motion is,
v = u + a t
Substitute the required value in the above expression.
0 = ( − 29.4 m / s ) + ( 9.8 m / s 2 ) t t = 3 s
The time of ascent and the time of descent are same, so the total time taken by the ball to return to the players is,
T = 2 t = 2 × 3 s = 6 s
Hence, the ball return to the player’s hand after 6 s.
Given, the initial speed of the ball is
(a)
When the ball is thrown vertically upwards, the only external force acting on the ball is gravitational force. Since the direction of action of gravitational force is in downward direction; therefore, the acceleration of the motion of the ball will be in the downward direction.
(b)
When the ball reaches its highest point the velocity of the ball at that instant is zero, and the acceleration of the ball is equal to the acceleration due to gravity, that is
(c)
Let the position of the ball at the highest point be
When the ball is in upward motion the sign of position and the acceleration of the ball is positive and the sign of the velocity of the ball is negative.
When the ball is in downward motion, the sign of position, velocity and the acceleration of the ball are positive.
(d)
Let
The third equation of motion is,
Substitute the required values in the above equation.
Thus, the ball rises to a height of 44.1 m.
The first equation of motion is,
Substitute the required value in the above expression.
The time of ascent and the time of descent are same, so the total time taken by the ball to return to the players is,
Hence, the ball return to the player’s hand after 6 s.
