A ball is thrown straight upward with a velocity and in a time reaches the top of its flight path, which is a displacement above the launch point. With a launch velocity of , what would be the time required to reach the top of its flight path and what would be the displacement of the top point above the launch point

i think the answer is a, hope this is right, that would be my best guess.

the minimum constant speed at which she can catch the train is 2.7 m/s

given:

acceleration = 0.6

time = 4.5 s

to find:

velocity = ?

formula used:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

solution:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

v = 0 + (0.6 × 4.5)

v = 2.7 m/s

the minimum constant speed at which she can catch the train is 2.7 m/s

blue light is scattered in different directions by the tiny molecules of air. blue is scattered more than other colours because it has a shorter distance to travel which results in smaller waves. this is why we see a blue sky most of the time.