# A pebble of mass 0.05 kg is thrown vertically upwards

A pebble of mass 0.05 kg is thrown vertically upwards. Give the direction and magnitude of the net force on the pebble .
(i) during its upward motion.
(ii) during its downward motion.
(iii) at the highest point where it is momentarily at rest.
Do your answer change if the pebble was thrown at an angle of 45° with the horizontal direction? Ignore air resistance.

When an object is thrown vertically upward or it falls vertically downward under gravity then an acceleration g = 10 m/\${{s}^{2}}\$ acts downward due to the earth’s gravitational pull.
Mass of pebble (m) = 0.05 kg
(i) During upward motion
Net force acting on pebble (F) = ma = 0.05 X10 N
= 0.50N (vertically downward)
(ii) During downward motion
Net force acting on pebble (F) = ma = 0.05 X10 N
= 0.50N (vertically downward)
(iii) At the highest point
Net force acting on pebble
(F) = ma = 0.05 X10N
= 0.50 N (vertically downward) If pebble was thrown at an angle of 45° with the horizontal direction then acceleration acting on it and therefore force acting on it will remain unchanged, i.e., 0.50 N (vertically downward). In case ©, at the highest point the vertical component of velocity will be zero but horizontal component of velocity will not be zero.