Asilver dollar is dropped from the top of a building that is 1344 feet tall. use the position function below for free-falling objects. s(t) = -16t^(2) + v0t + s0

(a) determine the position and velocity functions for the coin. s(t)= v(t)=

(b) determine the average velocity on the interval [3,4]. = ft/s

(c) find the instantaneous velocities when t = 3 s and t = 4 s. v(3)= v(4)=

(d) find the time required for the coin to reach the ground level. t=

(e) find the velocity of the coin at impact. vf =/s

a. position function of the coin:

    Now the velocity function:

b.

c.        &  

d.

e.

Explanation:

Given:

Given position function associated with free falling objects:

here:

initial height

initial velocity

time of observation

a)

position function of the coin:

∵the object is dropped it was initially at rest

Now the velocity function:

b)

we know average velocity is given as:

Displacement in the given interval:

Now,

c)

Instantaneous velocity at t = 3 s:

Instantaneous velocity at t = 4 s:

d)

At ground we  have s=0:

Put this in position function:

e)

Velocity of the coin at impact:

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.