SPEED

[Ryan Rigdon]

my problem

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?

Work so far, but I am a little lost with this problem.

dv/dt = -5

v = integral -5 dt = -5t + C

v = ds/dt -5t

s = - integral 5t dt

s = (-5t^2)/2

Confused?

 

[soroban]

Hello, Ryan Rigdon!

Is this a trick question?

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.

13 seconds later, Colleen is 65 feet above the ground.

Since the ball just makes it to Colleen, the maximum height of the ball is 65 feet.
The maximum height occurs at the vertex of the parabolic function.

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

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

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


$\displaystyle \text{We have: }\:\frac{v_o^2}{64} \:=\:65 \quad\Rightarrow\quad v_o \:=\:64\cdot65 \quad\Rightarrow\quad v \:=\:8\sqrt{65}$


$\displaystyle \text{Therefore, the ball should be thrown at: }\:64.49806199 \;\approx\;64.5\text{ ft/sec}$


~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

If the ball is thrown upward at 64.5 ft/second,
. . it takes roughly one second to reach a height of 65 feet.

But during that one second, the balloon has risen another 5 feet.
. . Hence, the ball will not reach Colleen.

It is necessary to thrown the ball at about 64.7 ft/sec.

 

[Ryan Rigdon]

not a trick question. it is exactly as it looks on my homework assignment.

 

[Ryan Rigdon]

the answer was 69.692

 

[Ryan Rigdon]

got this problem wrong so here it is again but with number change.

A hot-air balloon left the ground rising at 4 ft per second. 10 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?

 

[Ryan Rigdon]

here i go again

 

[Ryan Rigdon]

been trying to figure out how they got 69.692. if i figure this out i will be able to use this and do the problem again that i posted that use different numbers.

original problem

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?

their answer again 69.692

 

[Ryan Rigdon]

Hello, Ryan Rigdon!

Is this a trick question?

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.

13 seconds later, Colleen is 65 feet above the ground.

Since the ball just makes it to Colleen, the maximum height of the ball is 65 feet.
The maximum height occurs at the vertex of the parabolic function.

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

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

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


$\displaystyle \text{We have: }\:\frac{v_o^2}{64} \:=\:65 \quad\Rightarrow\quad v_o \:=\:64\cdot65 \quad\Rightarrow\quad v \:=\:8\sqrt{65}$


$\displaystyle \text{Therefore, the ball should be thrown at: }\:64.49806199 \;\approx\;64.5\text{ ft/sec}$


~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

If the ball is thrown upward at 64.5 ft/second,
. . it takes roughly one second to reach a height of 65 feet.

But during that one second, the balloon has risen another 5 feet.
. . Hence, the ball will not reach Colleen.

It is necessary to thrown the ball at about 64.7 ft/sec.

now i understand how they got 69.692. like previously stated it took one second to reach colleen but wont. and the reason is that the balloon rose another 5 feet hence again will not reach her. so for it to do that the ball should be thrown at 64.692 + 5 will give you 69.692 must be thrown initially to reach Colleen in time.

 

[Ryan Rigdon]

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 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?

 

[Ryan Rigdon]

quick from previously stated how can

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

wouldnt it be

$\displaystyle \text{Hence, maximum height is: }\:h \:=\:v_o\left(\frac{v_o}{32}\right) - 16\left(\frac{v_o}{32}\right)^2 \:=\:\frac{-15v_o^2}{32}\text{ feet.}$

 

[Ryan Rigdon]

Hello, Ryan Rigdon!

Is this a trick question?

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.

13 seconds later, Colleen is 65 feet above the ground.

Since the ball just makes it to Colleen, the maximum height of the ball is 65 feet.
The maximum height occurs at the vertex of the parabolic function.

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

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

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


$\displaystyle \text{We have: }\:\frac{v_o^2}{64} \:=\:65 \quad\Rightarrow\quad v_o \:=\:64\cdot65 \quad\Rightarrow\quad v \:=\:8\sqrt{65}$


$\displaystyle \text{Therefore, the ball should be thrown at: }\:64.49806199 \;\approx\;64.5\text{ ft/sec}$


~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

If the ball is thrown upward at 64.5 ft/second,
. . it takes roughly one second to reach a height of 65 feet.

But during that one second, the balloon has risen another 5 feet.
. . Hence, the ball will not reach Colleen.

It is necessary to thrown the ball at about 64.7 ft/sec.

how was this figured out soroban

It is necessary to thrown the ball at about 64.7 ft/sec.
[/size][/quote]

 

[Ryan Rigdon]

A hot-air balloon left the ground rising at 4 ft per second. 10 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.

10 seconds later, Colleen is 40 feet above the ground.

Since the ball just makes it to Colleen, the maximum height of the ball is 40 feet.
The maximum height occurs at the vertex of the parabolic function.

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

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

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


$\displaystyle \text{We have: }\:\frac{v_o^2}{64} \:=\:40 \quad\Rightarrow\quad v_o \:=\:64\cdot40 \quad\Rightarrow\quad v \:=\:8\sqrt{40}$


$\displaystyle \text{Therefore, the ball should be thrown at: }\:50.59644256 \;\approx\;50.596\text{ ft/sec}$

Needs to be rounded off at three decimal places.

Since it took about 1 second to get to Colleen it rose another 4 feet.

So in order to reach Colleen the ball must be thrown at a speed of 54.596 feet per second.

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~


NOTE: from the last problem 64.7 was used how was that figure out? 64.7 + 5 = 69.7. The answer to the original problem i started with was 69.692 feet per second in order to reach Colleen.

 

[Ryan Rigdon]

my answer was 54.596 got the problem wrong. the correct answer is 54.754.

what gives here please help.

 

[galactus]

Hi R. I ge the same as you.

$\displaystyle h=4t+40$

$\displaystyle v=-32t+v_{0}$

$\displaystyle h=-16t^{2}+v_{0}t$

When $\displaystyle -16t^{2}+v_{0}t=4t+40$

$\displaystyle v_{0}=4+16t+\frac{40}{t}$

$\displaystyle \frac{dv_{0}}{dt}=16-\frac{40}{t^{2}}$

$\displaystyle t=\frac{\sqrt{10}}{2}\approx 1.581 \;\ sec$

When she catches the ball, it is not at a high point but when the balloon is $\displaystyle 4(\frac{\sqrt{10}}{2})+40\approx 46.324 \;\ ft$

This means the initial velocity is $\displaystyle v_{0}=4+16(\frac{\sqrt{10}}{2})+\frac{40}{\frac{\sqrt{10}}{2}}\approx 54.596 \;\ ft/sec$

 

[Ryan Rigdon]

my answer was 54.596 got the problem wrong. the correct answer is 54.754.

what gives here please help.

i just dont understand how they ge their answer bugs me a lot. i know how to do the problem but simply for the love of god cant see where 54.754 came from. :evil:

 

[galactus]

It's possible it's a typo in the book?.

 

[Ryan Rigdon]

its actually an online quiz i got it from and it was not a typo.

 

[galactus]

As a matter of fact, a little while ago I found this exact problem and it agreed with our answer. That is what made me wonder about the answer you were given.

 

[Ryan Rigdon]

i emailed my professor telling him the situation with this problem. hopefully he will give me credit for it. did it twice and still got it wrong. BS if u ask me. i will post what he tells me.

 

[Deleted member 4993]

Galactus,

The speed you found is the minimum speed (d[sup:2atk6jz3]2[/sup:2atk6jz3]v/dt[sup:2atk6jz3]2[/sup:2atk6jz3] > 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:2atk6jz3]2[/sup:2atk6jz3]/64 and time of travel t = v/32

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

4*(10+v/32) = v[sup:2atk6jz3]2[/sup:2atk6jz3]/64

v[sup:2atk6jz3]2[/sup:2atk6jz3] - 8v - 2560 = 0 ? v = 54.75431016