Question

A player throws a ball upwards with an initial speed of $~29.4m{{s}^{-1}}$.
(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 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 = $9.8m/{{s}^{2}}$ and neglect air resistance).

Answer 1

Hint: To solve this question, we need a basic understanding of relative acceleration and velocity. Since, the ball is thrown upwards, the body has a negative gravitational acceleration or gravitational deceleration. Keeping this in mind, we can derive various formulas for motion of a body and can accordingly solve the given question.

Complete answer:
Let us answer the questions one by one:

a.)We should know that irrespective of the direction of the motion of the ball, acceleration (which is defined as the actual acceleration due to the gravity) always acts in the downward direction and is directed towards the centre of the Earth.

b.)We know that at the maximum height, velocity of the ball becomes zero. Therefore, the acceleration due to the gravity at a given place is constant and so it acts on the ball at all the points (considering the highest point) with a constant value. The value of the acceleration due to gravity is given as $9.8m/{{s}^{2}}$.

c.)We should know that during the upward motion, the sign of the position is positive. The sign of the velocity is considered to be negative and the sign of acceleration is positive. Now during the downward motion, the signs of position, velocity and acceleration are all considered to be positive.

d.)We know that it is given in the question that,
The initial velocity of the ball is u, which is given as $~29.4m{{s}^{-1}}$.
Now the final velocity of the ball or v is considered as 0, when the ball is at the maximum height and the velocity of the ball becomes zero.
As it is given in the question that the value of acceleration is $9.8m/{{s}^{2}}$.
So from the third law of motion, we can write that the height as:
${{v}^{2}}-{{u}^{2}}=2gs$
Now we can write that,
$s=({{v}^{2}}-{{u}^{2}})/2g$
Now putting the values in the above equation we can say that
$(({{0}^{2}})-{{(29.4)}^{2}})/2(-9.8)=3s$
Now we can write that
The time of ascent = The time of descent
Hence we can say that the total time taken by the ball to return to the player’s hands will be $\text{3s + 3s = 6s}$

Note:
We need to have clarity on the vectors because changes in direction can change the overall polarity of the vector fields of the body changing the resultant entirely. Keeping this in mind, we can carefully solve any such questions without making any silly mistakes.