Arock is thrown vertically upward from the ground, with an initial velocity of 64 ft/sec. its position function is s(t) = –16t2 + 64t. what is its velocity in ft/sec when it hits the ground?

Explanation:

Its position function is s(t) = –16t² + 64t

When it reaches back ground s(t) = 0 ft

Substituting

                s(t) = –16t² + 64t = 0

                          t² - 4t = 0

                           t²  = 4t

                           t =4 seconds

Now we need to find velocity when it reaches ground, that is velocity after 4 seconds.

Differentiating s(t) equation

                   v(t) = -32t+64

            Substituting t = 4 seconds

                  v(4) = -32 x 4+64  = -64 ft/s

Velocity of rock when it hits the ground is -64 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.

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