Question

A silver 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=___________s

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

Answer 1

Answer:

a. position function of the coin:

    $s=-16t^2+1344$

    Now the velocity function:

    $v=-32t$

b. $v_{avg}=-112\ m.s^{-1}$

c. $v_3=-96\ m.s^{-1}$        &   $v_4=-128\ m.s^{-1}$

d. $t=9.165\ s$

e. $v_f=293.285\ m.s^{-1}$

Explanation:

Given:

• height of dropping the silver dollar, $h=1344\ ft$

Given position function associated with free falling objects:

$s=-16t^2+v_0t+s_0$

here:

$s_0=$ initial height

$v_0=$initial velocity

$t=$ time of observation

a)

position function of the coin:

$s=-16t^2+1344$

∵the object is dropped it was initially at rest

Now the velocity function:

$v=\frac{d}{dt} s$

$v=-32t$

b)

we know average velocity is given as:

$\rm v_{avg}=\frac{total\ displacement}{total\ time}$

Displacement in the given interval:

$s_{_{3-4}}=s_4-s_3$

$s_{_{3-4}}=(-16\times 4^2+1344)-(-16\times 3^2+1344)$

$s_{_{3-4}}=-112\ ft$

Now,

$v_{avg}=\frac{-112}{4-3}$

$v_{avg}=-112\ m.s^{-1}$

c)

Instantaneous velocity at t = 3 s:

$v_3=-32\times 3$

$v_3=-96\ m.s^{-1}$

Instantaneous velocity at t = 4 s:

$v_4=-32\times 4$

$v_4=-128\ m.s^{-1}$

d)

At ground we  have s=0:

Put this in position function:

$0=-16t^2+1344$

$t=9.165\ s$

e)

Velocity of the coin at impact:

$v_f=-32\times 9.165$

$v_f=293.285\ m.s^{-1}$