A ball of mass $2 \mathrm{~kg}$ is thrown up with a velocity of $20 \mathrm{~ms^{-1}}$.

(A) Identify the sign of the work done by gravity on the ball during its upward motion and its downward motion.

(B) If the ball reaches a height of 19.4 m, how much work was done by air resistance (assume $g = 10 \text{ ms}^{-2}$).

(A) During the upward motion, the work done by gravity is negative because gravity acts downward while the ball moves upward. During the downward motion, the work done by gravity is positive because both gravity and displacement are downward.

(B) Given:

Mass of ball, $m = 2 \text{ kg}$

Initial velocity, $u = 20 \text{ ms}^{-1}$

Height reached, $h = 19.4 \text{ m}$

$$g = 10 \text{ ms}^{-2}$$

Initial kinetic energy of the ball,

$$K.E. = \frac{1}{2}mv^2$$

$$= \frac{1}{2} \times 2 \times (20)^2$$

$$= 400 \text{ J}$$

Potential energy at maximum height:

P.E. = $mgh$

$$= 2 \times 10 \times 19.4$$

$$= 388 \text{ J}$$

Work done by air resistance:

$$= 388 - 400$$

$$= -12 \text{ J}$$

Therefore, the work done by air resistance is -12 J.

The negative sign shows that air resistance opposes the motion of the ball.