A player throws a ball upwards with an initial speed of $29.4 \mathrm{~m} \mathrm{~s}^{-1}$.

A player throws a ball upwards with an initial speed of $29.4 \mathrm{~m} \mathrm{~s}^{-1}$.

(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 $x=0 \mathrm{~m}$ and $t=0 \mathrm{~s}$ 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 $g=9.8 \mathrm{~m} \mathrm{~s}^{-2}$ and neglect air resistance).



(a) Downward

(b) Velocity $=0$, acceleration $=9.8 \mathrm{~m} / \mathrm{s}^{2}$

(c) $x>0$ for both up and down motions, $v<0$ for up and $v>0$ for down motion, $a>0$ throughout the motion

(d) $44.1 \mathrm{~m}, 6 \mathrm{~s}$


(a) 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.

(b) 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., $9.8 \mathrm{~m} / \mathrm{s}^{2}$.

(c) 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.

(d) Initial velocity of the ball, $u=29.4 \mathrm{~m} / \mathrm{s}$

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

Acceleration, $a=-g=-9.8 \mathrm{~m} / \mathrm{s}^{2}$

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

$v^{2}-u^{2}=2 g s$

$s=\frac{v^{2}-u^{2}}{2 g}$

$=\frac{(0)^{2}-(29.4)^{2}}{2 \times(-9.8)}=44.1 \mathrm{~m}$

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

$v=u+a t$

$t=\frac{v-u}{a}=\frac{-29.4}{-9.8}=3 \mathrm{~s}$

Time of ascent $=$ Time of descentv

Hence, the total time taken by the ball to return to the player's hands $=3+3=6 \mathrm{~s}$.

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