$A$ pebble of mass $0.05 \, kg$ is thrown vertically upwards. Give the direction and magnitude of the net force on the pebble,$(a)$ during its upward motion,$(b)$ during its downward motion,$(c)$ at the highest point where it is momentarily at rest. Do your answers change if the pebble was thrown at an angle of $45^{\circ}$ with the horizontal direction? Ignore air resistance.

(A) The net force is $0.5 \, N$ in the vertically downward direction in all cases.
Acceleration due to gravity $(g)$,irrespective of the direction of motion of an object,always acts downward. The gravitational force is the only force acting on the pebble in all three cases. Its magnitude is given by Newton's second law of motion as:
$F = m \times a$
Where,$F$ is the net force,$m$ is the mass of the pebble $(0.05 \, kg)$,and $a = g = 10 \, m/s^2$.
Therefore,$F = 0.05 \times 10 = 0.5 \, N$.
The net force on the pebble in all three cases is $0.5 \, N$ and this force acts in the downward direction.
If the pebble is thrown at an angle of $45^{\circ}$ with the horizontal,it will have both horizontal and vertical components of velocity. At the highest point,only the vertical component of velocity becomes zero. However,the pebble will have a horizontal component of velocity throughout its motion. This component of velocity has no effect on the net force acting on the pebble,which remains $0.5 \, N$ downwards.