If a ball is thrown vertically upwards with a velocity of 40 m/s, then what will be the magnitude of its displacement after 6 s? (Take g = 10 m/s$$^2$$)

Ball is thrown vertically upward with velocity 40m/s . at highest point, velocity of ball equals zero.

use formula, v = u + at,
here, v = 0, u = 40 m/s and a = -g [ negative sign shows that acceleration due to gravity acts downward direction.]
so, 0 = 40 - 10t
t = 4sec
and maximum height reached by ball = $$\frac{u^2}{2g}$$
=$$\frac{40^2}{20}$$= 80m
hence, after 4sec ball reaches highest position.
then, ball starts to fall downward direction.
so, initial velocity in this case , u= 0
after 2 sec ball falls , S distance below from highest point.
use formula, S = ut + $$\frac{1}{2}at^2$$
S = -20m

hence,first 4sec ball reaches 80m heigh from the ground then, falls 20m below from the top point in next 2 sec .so, displacement in 6sec = 80m - 20m = 60m