A pebble of mass 0.05 kg is thrown vertically upwards Give the magnitude and direction of net force on the pebble (a) during its upward motion (b) during its downward motion (c) at the highest point where it is momentarily rest
Do your answers change if the pebble were thrown at an angle of say `45^(@)` horizontal direction lgnore air resistance .

Explanation

(a) When the pebble is moving upward ,the acceleration g is acting downward , so the force is acting downward is equal to F = mg = `0.05 kg xx 10 ms^(-2)` = 0.5 N .
(b) In this case also F = mg = `0.05 xx 10 ` = 0.5 N (downwards)
(c ) The pebble is not at rest at highest point but has horizontal component of velocity. The direction and magnitude of the net force on the pebble will not alter even if it thrown at `45^(@)` because no other acceleration except 'g' is acting on pebble.