Question

if a ball is thrown vertically upward with velocity 40 m/sec.,then after 6 second what will be magnitude of displacement?

Answer 1

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 = u²/2g
= (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 + 1/2at²
S = 0.t + 1/2(-g) × 2² = -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

and distance = 80m + 20m = 100m

Answer 2

A ball is thrown in the upward direction with the initial velocity (u) of 40 m/sec. since the ball is going upward hence acceleration will be gravitational acceleration that is 10m/sec.

On reaching he top, the final velocity (v) will be zero at the point it started to come down,

v = u + at

0 = 40 * (-10) * t

10t = 40

t = 4 sec

hence the maximum height of the ball will be,

2gs = v^2 – u^2

s = v^2/2g

s = 40^2/2 * 10 = 1600/20 = 80 m

after the completion of 4 sec, ball will start to move in the downward direction with u = 0 m/s

hence, s = ut + ½ gt^2

s = 0 + ½ * -10 * 2^2

s = ½ * -10 * 4

s = -20 m

hence, the displacement = 80 m – 20 m = 60 m