Question

 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° with the horizontal direction?
Ignore air resistance.

Answer 1

I think that ,
(a)during its upward motion ,
its direction will be upward and force will be f=ma =>f=0.05×(-9.8)= -0.49N
(b)during its downward motion,
its direction will be downward and force will be f=ma =>f=0.05×9.8=0.49N
(c)at the highest point where it is momentarily at rest ,
the net force will be 0 and it will have no direction at that point because the force is 0.
Yes ,the answers will change because its direction will change .

Answer 2

Answer:

(a)during its upward motion ,

its direction will be upward and force will be f=ma =>f=0.05×(-9.8)= -0.49N

(b)during its downward motion,

its direction will be downward and force will be f=ma =>f=0.05×9.8=0.49N

(c)at the highest point where it is momentarily at rest ,

the net force will be 0 and it will have no direction at that point because the force is 0.

Yes ,the answers will change because its direction will change .