If a ball is thrown vertically upward with a velocity of 80 ft/s, then its height after seconds is =−. (a) what is the maximum height reached by the ball? (b) what is the velocity of the ball when it is 96 ft above the ground on its way up? (c) what is the velocity of the ball when it is 96 ft above the ground on its way down? if a ball is thrown vertically upward with a velocity of 80 ft/s, then its height after seconds is =−. (a) what is the maximum height reached by the ball? (b) what is the velocity of the ball when it is 96 ft above the ground on its way up? (c) what is the velocity of the ball when it is 96 ft above the ground on its way down?

100 ft

16 ft/s

16 ft/s

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 32 ft/s² (negative upwards and positive downwards)

The function here is of the form which is one of the equations of motion

From another equation of motion

The maximum height the ball will reach is 100 ft

When s = 96 ft

The velocity of the ball when it is 96 ft above the ground on its way up is 16 ft/s

On the way down when it will be 96 ft above the ground it would travel 100-96 = 4 ft from the maximum height

The velocity of the ball on the way down when it is 96 ft above the ground  is 16 ft/s

the box does not move because you need to use a force greater than 300 n. you are only using 50 n. it needs to at least be 301 n for the box to move because then it is an unbalanced force and it will move the way that the greater force is pushing it. if you use only 300 n it will be a balanced force and will not move because the force you are exerting on the   box is the same as its force.

hope this !

im sorry if i went too in depth, i am going to include a link to a video on khan academy that will explain this.   hope that you like the video and that it explains more in detail to you.

that is the link.