SPEED

[soroban]

Hello, Ryan Rigdon!

There was a typo in my second answer . . . sorry!

A hot-air balloon left the ground rising at 5 ft per second.
13 seconds later, Victoria through a ball straight up to her friend Colleen in the balloon.
At what speed did she throw the ball if it just made it to Colleen?

The height of the ball is given by: .$\displaystyle h \:=\:v_ot - 16t^2$
. . where $\displaystyle v_o$ is the initial speed of the ball.

The maximum height occurs at the vertex of the parabolic function.

$\displaystyle \text{The vertex is at: }\:t = \frac{\text{-}b}{2a},\,\text{ where }b = v_o\,\text{ and }\,a = \text{-}16$

$\displaystyle \text{Maximum height occurs at: }\:t \:=\:\frac{\text{-}v_o}{2(\text{-}16)} \:=\:\frac{v_o}{32}\text{ seconds.}$

$\displaystyle \text{Hence, maximum height is: }\:h \:=\:v_o\left(\frac{v_o}{32}\right) - 16\left(\frac{v_o}{32}\right)^2 \:=\:\frac{v_o^2}{64}\text{ feet. }\;[1]$


$\displaystyle \text{Colleen was already 65 feet above the ground.}$

$\displaystyle \text{When the ball reaches her }\left(\tfrac{v_o}{32}\text{ seconds later}\right)\!,\text{ she has risen another }5\!\left(\tfrac{v_o}{32}\right)\text{ feet.}$

$\displaystyle \text{Hence, Colleen's height is: }\:65 + \frac{v_o}{32}\text{ feet. }\;[2]$


$\displaystyle \text{Equate [1] and [2]: }\;\frac{v_o^2}{64} \;=\;65 + \frac{5v_o}{32} \quad\Rightarrow\quad v_o^2 - 10v_o - 4160 \:=\:0$

$\displaystyle \text{Quadratic Formua: }\;v_o \;=\;\frac{-(\text{-}10) \pm\sqrt{(\text{-}10)^2 - (4)(1)(\text{-}4160)}}{2} \;=\;\frac{10 \pm\sqrt{16740}}{2} \;=\;5\;\pm\; 3\sqrt{465}$


$\displaystyle \text{Therefore: }\:v_o \;=\;69.69157596 \;\approx\;\boxed{69.7\text{ ft/sec.}}$

 

[galactus]

Galactus,

The speed you found is the minimum speed (d[sup:1krfft00]2[/sup:1krfft00]v/dt[sup:1krfft00]2[/sup:1krfft00] > 0)required for the ball to be caught.

If the ball is "just" caught - we ned to assume that the greatest height of the balls trajectory touches the baloon.

h = v[sup:1krfft00]2[/sup:1krfft00]/64 and time of travel t = v/32

Then baloon will travel 4*(10+v/32). So

4*(10+v/32) = v[sup:1krfft00]2[/sup:1krfft00]/64

v[sup:1krfft00]2[/sup:1krfft00] - 8v - 2560 = 0 ? v = 54.75431016

I misunderstood the question then. I see now. I reckon I did not read the "just made it to"

 

[Ryan Rigdon]

thanx galactus. i am hoping the professor will let me take the quiz again. i will post my findings as soon as i find out.

 

[Ryan Rigdon]

A hot-air balloon left the ground rising at 6 ft per second.
12 seconds later, Victoria through a ball straight up to her friend Colleen in the balloon.
At what speed did she throw the ball if it just made it to Colleen?

Height of the ball $\displaystyle h \:=\:v_ot - 16t^2$
. . where $\displaystyle v_o$ is the initial speed of the ball.

The maximum height occurs at the vertex of the parabolic function.

$\displaystyle \text{The vertex is at: }\:t = \frac{\text{-}b}{2a},\,\text{ where }b = v_o\,\text{ and }\,a = \text{-}16$

$\displaystyle \text{Maximum height occurs at: }\:t \:=\:\frac{\text{-}v_o}{2(\text{-}16)} \:=\:\frac{v_o}{32}\text{ seconds.}$

$\displaystyle \text{Hence, maximum height is: }\:h \:=\:v_o\left(\frac{v_o}{32}\right) - 16\left(\frac{v_o}{32}\right)^2 \:=\:\frac{v_o^2}{64}\text{ feet. }\;[1]$


$\displaystyle \text{Colleen was already 72 feet above the ground.}$

$\displaystyle \text{When the ball reaches her }\left(\tfrac{v_o}{32}\text{ seconds later}\right)\!,\text{ she has risen another }6\!\left(\tfrac{v_o}{32}\right)\text{ feet.}$

$\displaystyle \text{Hence, Colleen's height is: }\:72 + \frac{v_o}{32}\text{ feet. }\;[2]$


$\displaystyle \text{Equate [1] and [2]: }\;\frac{v_o^2}{64} \;=\;72 + \frac{5v_o}{32} \quad\Rightarrow\quad v_o^2 - 10v_o - 4160 \:=\:0$

\(\displaystyle \text{Quadratic Formua: }\;v_o \;=\;\frac{-(\text{-}12) \pm\sqrt{(\text{-}12)^2 - (4)(1)(\text{-}4608)}}{2} \;=\;\frac{12 \pm\sqrt{18576}}{2} \\)


$\displaystyle \text{Therefore: }\:v_o \;=\;74.14690015 \;\approx\;\boxed{74.147\text{ ft/sec.}}$

[/size][/quote]

had to be rounded off to three decimal places. got it right finally. thanx for all the help.

 

[Deleted member 4993]

Ryan,

The critical point in these problems is the wording. When it says "just reaches the baloon" we are assuming that the the ball reaches the baloon at its maximum height.

But that is not the minimum speed at which the ball can be thrown - and still reach the baloon. The way Galactus (and you) had solved the problem - that is what you had calculated.

 

[TchrWill]

Here is my problem same type with different number will report back with my work shortly.

A hot-air balloon left the ground rising at 4 ft per second. 10 seconds later, Victoria threw a ball straight up to her friend Colleen in the balloon. At what speed did she throw the ball if it just made it to Colleen?

Just another viewpoint.

1--Assume Colleen's hands reach 4x10 = 40 feet altitude in 10 seconds rising at 4 ft./sec.
2--At this instant, Victoria throws the ball upward with a speed of Vo ft./sec. (Both the ball and Colleen's hands are now in motion).
3--The height of the ball is defined by h = Vo(t) - 16(t^2) = (40 + d) where d = the distance from the height of 40 ft. to the ultimate height where Colleen's hands and the ball meet.
4--The speed of the ball is defined by Vf = Vo - 32(t) where Vf = the ball's speed when it reaches Colleen's hands.
5--With Vf = "0", Vo = 32(t) or t = Vo/32 (the ball just made it to Colleen).
6--Substituting into (3) yields t = sqrt[40 + d)/16].
7--In this same period of time, t, the balloon rises from the 40 foot altitude to the point where Colleens hands and the ball meet, or t = d/4.
8--Therefore, d/4 = sqrt[(40 + d)/16] which simplifies to d^2 - d - 40 = 0.
9--This results in d = 6.844 feet yielding h = 46.844 feet, t = 1.711 seconds and an initial speed of the ball of 54.75 ft./sec.