Read the Details below and show all of your work!?

Question

The path of a falling object is given by the function s = 16t² + v_0t + s_0 where represents the initial velocity in ft/sec and s_0 represents the initial height. The variable t is time in seconds, and s is the height of the object in feet.

IF a rock is thrown upward with an initial velocity of 32 feet per second from the top of a 40-foot building, write the height equation using this information.

Typing hint: Type t-squared as t^2

Answer 1

The initial equation is given as a FALLING object, so positive values are downward:

s = 16t² + v_0t + s_0

The height of the object is:

-H = 16t² - 32t - 40

v_0 is given as 32 and

s_0 is given as 40.

The signs are negative for H, V_0 and s_0 because these values are in the opposite direction. Rearranging for convenience gives:

H = -16t² + 32t + 40

At 1/2 second the height is then:

H_0.5 = -4+16+40

= 52ft

To find the maximum value, first find the first derivative of the equation:

H' = -32t +32

Set it to zero and solve for time

0 = -32t +32

t=1 second

At 1 second the height will be

H = -16t² + 32t + 40

= -16 + 32 + 40

56 ft

Answer 2

s(t) = 16t² + v_0t + s_0

s(t) = 16t² + 32t + 40

s(.5) = 16(5²) + 32(5) + 40 = 600.

ds/dt = 32t +32=0 does not have a positive root for t.

Something is wrong; are you sure the original equation is right?

Answer 3

These are fun...

To learn more about these kind of equations look up

"Equations of Motion" or "Kinematics"

This function

s = 16t² + v_0t + s_0

comes from the general function for a projectile...

X - X_0 = V_0 t + 1/2 a t²

Where

X - X_0  = change in position

V_0 = Initial Velocity

a = acceleration due to gravity  -9.8 m/s² or -32 ft/s²

By the way your initial equation for position is wrong.

Here is the corrected general equation for

position as a function of time

s - s_0  = v_0 t + 1/2 * -32 t²

Notice your accelearation is negative.

This is because objects accelerate in the downward direction which is typically negative...

Anyhow our equation simplifies to...

s = v_0 t - 16 t² + s_0

This means we start out a an initial posion, a rooftop or or ground floor s_0.  Then we launch the object into the air at an initial velocity v_0.

Thus as time increases it causes the position to change at a quadratic rate...

s(t) = v_0 t - 16 t² + s_0

This is the basic physics mechanic behind projectile motion.

Now back to your question...

Lets first rewrite the general equation to meet your initial conditions...

v_0 = 32 ft/s

s_0 = 40 ft

s(t) = 32 t - 16 t² + 40

First lets find the position of the object 0.5s into the throw...

s(0.5) = 32 (0.5) - 16 (0.5)² + 40

s(0.5) = 16 - 4 + 40 = 52

So after 0.5s the ball is at a height of 52 ft. from the ground.

In otherwords, from the roof, the objectl has traveled 12 ft upwards.

--------------------------------------...

For the next part, there are several ways to do it depending on how you want to think about it....  I will briefly show you each.

Objectives: Find Maximum height, and time at max

Imagine the motion of the object as a parabola.

( an upside down u)

Here is the math point of view

The top of the arc is what is called a critical point.

Calculus would find this using the derivative.

The derivative of position with respect to time is velocity.

s(t) = 32 t - 16 t² + 40

v(t) = ds /dt = 32 t - 32

To find the critical point we solve for when the derivative is 0.

0 = 32 t - 32

32 = 32 t

t = 1.0 s

So after 1s the max height is reached.

Now we solve for s(1) to get our max height.

s(1) = 32  - 16  + 40 = 56

So our max height is 56 ft.

--------------------------------------...

Now we look at it from a physics point of view...

The 3rd Equation of Kinematic Motion which relates final velocity and position.

V² - V_0² = 2 a (X - X_0)

We have an initial velocity

V_0 = 32 ft/s

We know the acceleration due to gravity is

a = -32 ft/s²

Lastly we know at the peak of the arc, the final velocity is

V = 0 ft/s

We can solve for the change in position.

Note, this is the maximum height with respect to the roof.

To get the actual height we have to add 40 ft.

For that reason we include X_0 to our equation.

Since we are solving for position we solve for the X-X_0

V² - V_0² = 2 a (X - X_0)

(X - X_0) = (V² - V_0²)  /  2 a

X = (V² - V_0²)  /  2 a  + X_0

Plug in our knowns...

X = (0² - 32²)  /  2 (-32)  + 40

X = - (32)(32) / 2 ( -32) + 40

X = 32 / 2 + 40

X = 16 + 40

X = 56

Again our result is the same, 56 ft.

We then use that max height to solve for the time

56 = 32 t - 16 t² + 40

To solve this we have to use the quadratic equation.

I know, now your really excited about this.. ☻

0 = 32 t - 16 t² + 40 - 56

0 = 32 t - 16 t²  -16

0 = - 16 t²  + 32 t -16

x =

- b ± √ b² - 4 a c

------------------------

2a

where

a = -16

b = 32

c = -16

-32 ± √(  (-32)² -  4*(-16)*(-16) )

--------------------------------------...

2*(-16)

-32 ± √(1024 - 1024)

-----------------------------

-32

The radical cancels itself out...

-32

-----

-32

= 1

So we have t = 1.0s

Personally I like the math way alot better...