Two Balls Thrown Vertical Upward

[mathdad]

If an object is thrown vertically upward from a height of h_0 feet with an initial speed of v_0 feet per second, then its height H (in feet) after t seconds is given by

h = -16t^2 + vt + h

One ball is thrown vertically upward from a height of 50 feet with an initial speed of 40 ft/sec. At the same time, another ball is thrown vertically upward from a height of 100 feet with an initial speed of 5 ft/sec. Which ball hits the ground first?

Must I use the formula TWICE, once for each ball? We are looking for t in both cases, right?

Ball 1

h_0 = 50 feet

v_0 = 40 ft/sec

We must find t in seconds.

What is h? Is h also 50 feet?

Ball 2

h_0 = 100 feet

v_0 = 5 ft/sec

We must find t in seconds.

What is h for ball 2? Is h also 100 feet?

The ball to hit the ground first is the ball traveling faster, which means the ball with the smaller t value in terms of seconds.

Correct?

 

[MarkFL]

If an object is thrown vertically upward from a height of h_0 feet with an initial speed of v_0 feet per second, then its height H (in feet) after t seconds is given by

h = -16t^2 + vt + h

I would use the following notation:

\(\displaystyle h(t)=-16t^2+v_0t+h_0\)

This way, it is more clear that the two parameters with a 0 subscript are initial values, and not instantaneous. I see you do use this notation later.

One ball is thrown vertically upward from a height of 50 feet with an initial speed of 40 ft/sec. At the same time, another ball is thrown vertically upward from a height of 100 feet with an initial speed of 5 ft/sec. Which ball hits the ground first?

Must I use the formula TWICE, once for each ball? We are looking for t in both cases, right?

Yes, you need to use the parameters (initial velocity and height) for each ball, and then observe that when a ball hits the ground, we have:

\(\displaystyle h(t)=0\)

This will give you two quadratics to solve, and you are interested in the positive roots only.

...The ball to hit the ground first is the ball traveling faster, which means the ball with the smaller t value in terms of seconds.

Correct?

The ball that hits the ground first isn't necessarily the one that has the greater speed when hitting the ground, it is simply the ball that takes the lesser time to hit the ground. :)

 

[MarkFL]

After a bit of time to think, I realize what I would do is begin with:

\(\displaystyle -16t^2+v_0t+h_0=0\)

Solve for $t$ using the quadratic formula, and discard the negative root:

\(\displaystyle t=\frac{v_0+\sqrt{v_0^2+64h_0}}{32}\)

This way we needed to solve only 1 quadratic, and we now have a formula in which to plug the respective parameters. ;)

 

[mathdad]

After a bit of time to think, I realize what I would do is begin with:

\(\displaystyle -16t^2+v_0t+h_0=0\)

Solve for $t$ using the quadratic formula, and discard the negative root:

\(\displaystyle t=\frac{v_0+\sqrt{v_0^2+64h_0}}{32}\)

This way we needed to solve only 1 quadratic, and we now have a formula in which to plug the respective parameters. ;)

Yes, it makes more sense to use the quadratic formula you provided.

1. Do I use the formula twice, once for each ball?

2. Where did 64 and 32 come from?

 

[MarkFL]

Yes, it makes more sense to use the quadratic formula you provided.

1. Do I use the formula twice, once for each ball?

Yes, for the first ball, we have:

\(\displaystyle v_0=40,\,h_0=50\)

And for the second ball we have:

\(\displaystyle v_0=5,\,h_0=100\)

2. Where did 64 and 32 come from?

They came from the application of the quadratic formula...

\(\displaystyle t=\frac{-v_0-\sqrt{v_0^2-4(-16)(h_0)}}{2(-16)}=\frac{v_0+\sqrt{v_0^2+64h_0}}{32}\)

 

[mathdad]

Yes, for the first ball, we have:

\(\displaystyle v_0=40,\,h_0=50\)

And for the second ball we have:

\(\displaystyle v_0=5,\,h_0=100\)
They came from the application of the quadratic formula...

\(\displaystyle t=\frac{-v_0-\sqrt{v_0^2-4(-16)(h_0)}}{2(-16)}=\frac{v_0+\sqrt{v_0^2+64h_0}}{32}\)

The quadratic you provided makes the problem so much easier. I thank you for your help. Can the formula you provided be used for similar questions involving objects that are thrown vertically upward?

 

[MarkFL]

The quadratic you provided makes the problem so much easier. I thank you for your help. Can the formula you provided be used for similar questions involving objects that are thrown vertically upward?

This formula will work for the given quadratic, which results from using an acceleration due to gravity of 32 ft/s². This is approximately the acceleration due to gravity at the Earth's surface. You would need to use another value if the balls were being thrown on another celestial body.

Note that this model neglects the effects of drag (air resistance) as well as other effects. As you progress in your studies, you will encounter more complex, but accurate, models. :)

 

[mathdad]

This formula will work for the given quadratic, which results from using an acceleration due to gravity of 32 ft/s². This is approximately the acceleration due to gravity at the Earth's surface. You would need to use another value if the balls were being thrown on another celestial body.

Note that this model neglects the effects of drag (air resistance) as well as other effects. As you progress in your studies, you will encounter more complex, but accurate, models. :)

1. What would be the formula for an object thrown vertically downward from a given height h?

2. Wind resistance is more a physics problem, right?

 

[MarkFL]

1. What would be the formula for an object thrown vertically downward from a given height h?

Since the initial velocity is downward, we would give it a negative sign...for example if the object is thrown downward at 5 m/s, we would use:

\(\displaystyle v_0=-5\,\frac{\text{m}}{\text{s}}\)

The way we have set up our coordinate axis in this problem, up is in the positive direction, and down is in the negative direction.

2. Wind resistance is more a physics problem, right?

All of kinematics (the study of bodies in motion) is part of physics, but they make good problems in the applications of various types of mathematics, because it is something with which we typically have experience in the real world and can easily visualize.

When we begin to consider drag, at least when I was a student, is in the study of linear first order differential equations. The first model I recall studying assumes that drag is proportional to velocity. Another model assumes drag is proportional to the square of velocity. These models assume that the coefficient of drag remains constant, which later models will address by using a model that accounts for the change in density of the atmosphere according to height. It can get as complicated as you care to make it. :)

 

[mathdad]

Since the initial velocity is downward, we would give it a negative sign...for example if the object is thrown downward at 5 m/s, we would use:

\(\displaystyle v_0=-5\,\frac{\text{m}}{\text{s}}\)

The way we have set up our coordinate axis in this problem, up is in the positive direction, and down is in the negative direction.
All of kinematics (the study of bodies in motion) is part of physics, but they make good problems in the applications of various types of mathematics, because it is something with which we typically have experience in the real world and can easily visualize.

When we begin to consider drag, at least when I was a student, is in the study of linear first order differential equations. The first model I recall studying assumes that drag is proportional to velocity. Another model assumes drag is proportional to the square of velocity. These models assume that the coefficient of drag remains constant, which later models will address by using a model that accounts for the change in density of the atmosphere according to height. It can get as complicated as you care to make it. :)

For objects thrown downward from a height h, would the quadratic formula be similar to the one you created in my sample question? I have no plans to study physics.

 

[MarkFL]

For objects thrown downward from a height h, would the quadratic formula be similar to the one you created in my sample question?

It would be identical (we are using the same model), we would just use a negative value for $v_0$, where the negative sign denotes the fact that the initial velocity is in the downward direction. :)

 

[mathdad]

Can you provide a sample question where an object is thrown vertically downward from a bridge, say, 150 feet in height?

 

[MarkFL]

Can you provide a sample question where an object is thrown vertically downward from a bridge, say, 150 feet in height?

Let's begin with a more general kinematic equation, that can model a body subject to one external force, such as gravity. Suppose the external force points in the negative direction, resulting in an acceleration $a$. Then it can be shown that the body's position $x$ as a function of time $t$ is given by:

\(\displaystyle x(t)=-\frac{1}{2}at^2+v_0t+x_0\)

At what times do we have:

\(\displaystyle x(t)=0\) ?

 

[mathdad]

Let's begin with a more general kinematic equation, that can model a body subject to one external force, such as gravity. Suppose the external force points in the negative direction, resulting in an acceleration $a$. Then it can be shown that the body's position $x$ as a function of time $t$ is given by:

\(\displaystyle x(t)=-\frac{1}{2}at^2+v_0t+x_0\)

At what times do we have:

\(\displaystyle x(t)=0\) ?

We set the given equation to 0 and solve for t, right?

 

[MarkFL]

We set the given equation to 0 and solve for t, right?

Yes. :)

 

[mathdad]

Let's begin with a more general kinematic equation, that can model a body subject to one external force, such as gravity. Suppose the external force points in the negative direction, resulting in an acceleration $a$. Then it can be shown that the body's position $x$ as a function of time $t$ is given by:

\(\displaystyle x(t)=-\frac{1}{2}at^2+v_0t+x_0\)

At what times do we have:

\(\displaystyle x(t)=0\) ?

How can I solve for t when a, v_0 and x_0 are not given?

 

[MarkFL]

How can I solve for t when a, v_0 and x_0 are not given?

You can still solve the equation, you just won't get numbers, you'll get expressions containing those parameters. We would begin by writing:

\(\displaystyle -\frac{1}{2}at^2+v_0t+x_0=0\)

I would begin by multiplying through by -2:

\(\displaystyle at^2-2v_0t-2x_0=0\)

Now we can apply the quadratic formula:

\(\displaystyle t=\frac{2v_0\pm\sqrt{4v_0^2-4(a)(-2x_0}}{2a}=\frac{2v_0\pm2\sqrt{v_0^2+2ax_0}}{2a}=\frac{v_0\pm\sqrt{v_0^2+2ax_0}}{a}\)

Suppose we define the distance traveled by the body as:

\(\displaystyle \Delta x=x(t)-x_0\)

Then our kinematic equation becomes:

\(\displaystyle -\frac{1}{2}at^2+v_0t-\Delta x=0\)

Can you solve for $t$ here?

 

[mathdad]

You can still solve the equation, you just won't get numbers, you'll get expressions containing those parameters. We would begin by writing:

\(\displaystyle -\frac{1}{2}at^2+v_0t+x_0=0\)

I would begin by multiplying through by -2:

\(\displaystyle at^2-2v_0t-2x_0=0\)

Now we can apply the quadratic formula:

\(\displaystyle t=\frac{2v_0\pm\sqrt{4v_0^2-4(a)(-2x_0}}{2a}=\frac{2v_0\pm2\sqrt{v_0^2+2ax_0}}{2a}=\frac{v_0\pm\sqrt{v_0^2+2ax_0}}{a}\)

Suppose we define the distance traveled by the body as:

\(\displaystyle \Delta x=x(t)-x_0\)

Then our kinematic equation becomes:

\(\displaystyle -\frac{1}{2}at^2+v_0t-\Delta x=0\)

Can you solve for $t$ here?

I can solve for t but it will lead to another expression. I will work on this from home later tonight.

 

[mathdad]

You can still solve the equation, you just won't get numbers, you'll get expressions containing those parameters. We would begin by writing:

\(\displaystyle -\frac{1}{2}at^2+v_0t+x_0=0\)

I would begin by multiplying through by -2:

\(\displaystyle at^2-2v_0t-2x_0=0\)

Now we can apply the quadratic formula:

\(\displaystyle t=\frac{2v_0\pm\sqrt{4v_0^2-4(a)(-2x_0}}{2a}=\frac{2v_0\pm2\sqrt{v_0^2+2ax_0}}{2a}=\frac{v_0\pm\sqrt{v_0^2+2ax_0}}{a}\)

Suppose we define the distance traveled by the body as:

\(\displaystyle \Delta x=x(t)-x_0\)

Then our kinematic equation becomes:

\(\displaystyle -\frac{1}{2}at^2+v_0t-\Delta x=0\)

Can you solve for $t$ here?

-(1/2)at^2 + v_0 t - ⟁x = 0

-(1/2)at^2 + v_0 t = ⟁x

t[-(1/2)at + v_0] = ⟁x

t = ⟁x/[(1/2)at + v_0]

This cannot be what solving for t is because there is a t on both sides.

 

[MarkFL]

-(1/2)at^2 + v_0 t - ⟁x = 0

-(1/2)at^2 + v_0 t = ⟁x

t[-(1/2)at + v_0] = ⟁x

t = ⟁x/[(1/2)at + v_0]

This cannot be what solving for t is because there is a t on both sides.

You have a quadratic equation...why not use the quadratic formula?

 

[mathdad]

You have a quadratic equation...why not use the quadratic formula?

-(1/2)at^2 + v_0 t - ⟁x = 0

t = -v_0 sqrt{(v_0)^2 - 4(-1/2)a(-⟁x)}/2(-1/2)a

t = -(v_0)•sqrt{(v_0)^2 - 2a⟁x}/(-a)

Yes?

 

[MarkFL]

-(1/2)at^2 + v_0 t - ⟁x = 0

t = -v_0 sqrt{(v_0)^2 - 4(-1/2)a(-⟁x)}/2(-1/2)a

t = -(v_0)•sqrt{(v_0)^2 - 2a⟁x}/(-a)

Yes?

No, that's not quite it...:)

 

[mathdad]

What is a, b and c for me to plug into the quadratic formula?

 

[MarkFL]

What is a, b and c for me to plug into the quadratic formula?

$a$ is the coefficient of the term containing $t^2$, $b$ is the coefficient of the term containing $t$, and $c$ is the constant term...

 

[mathdad]

$a$ is the coefficient of the term containing $t^2$, $b$ is the coefficient of the term containing $t$, and $c$ is the constant term...

In that case, a = -(1/2)a, b = v_0 abd c = ⟁x.

If not, then show me how this is done. I want to concentrate on solving word problems.

 

[MarkFL]

In that case, a = -(1/2)a, b = v_0 abd c = ⟁x.

If not, then show me how this is done. I want to concentrate on solving word problems.

It's kind of unfortunate that $a$ is being used for acceleration, and we also have $a$ in the quadratic formula. I will use $a'$ for the quadratic formula value.

We would have:

\(\displaystyle a'=-\frac{1}{2}a,\,b=v_0,=,c=-\Delta x\)

I know you wish to focus on word problems, but there is a method to my madness here, so please bear with me. It is important to be able to apply the theorems being studied, such as the quadratic formula. Once we have that done, then we can focus on looking at how the behavior of the parameters affects the roots. This was a skill I developed as a student, and it gave me better insight to the inner workings of the mathematics. ;)

 

[mathdad]

t = -v_0 ± sqrt{(v_0)^2 - 4(-1/2)a (- ⟁x)}/2(-1/2)a

t = -v_0 ± sqrt{v^2_0 + 2a (- ⟁x)}/-a

t = -v_0 ± sqrt{v^2_0 - 2a⟁x)}/-a

 

[MarkFL]

t = -v_0 ± sqrt{(v_0)^2 - 4(-1/2)a (- ⟁x)}/2(-1/2)a

t = -v_0 ± sqrt{v^2_0 + 2a (- ⟁x)}/-a

t = -v_0 ± sqrt{v^2_0 - 2a⟁x)}/-a

Another issue we need to address here and now is getting you to use $\LaTeX$. What you posted is hard to read for these poor old eyes...(Crying)

$\LaTeX$ needs to be wrapped in tags...the easiest to use are the MATH tags, which can be generated by clicking the $\Sigma$ button on our toolbar. Once you have those tags created, your cursor will be located within them, ready to accept your code. You want the code:

t=\frac{-v_0\pm\sqrt{v_0^2-4\left(-\frac{1}{2}a\right)\left(-\Delta x\right)}}{2\left(-\frac{1}{2}a\right)}=\frac{v_0\pm\sqrt{v_0^2-2a\Delta x}}{a}

This, within the MATH tags, produces:

\(\displaystyle t=\frac{-v_0\pm\sqrt{v_0^2-4\left(-\frac{1}{2}a\right)\left(-\Delta x\right)}}{2\left(-\frac{1}{2}a\right)}=\frac{v_0\pm\sqrt{v_0^2-2a\Delta x}}{a}\)

Take a bit of time to see how this works. You will find that just about every symbol/command you will ever need is located on our Quick $\LaTeX$ element to help you construct your equations/expressions. (Yes)

Why am I pushing you to learn this? You will find people will be more willing to jump in and help when they can more easily read your work. It just looks better, and is more informative for everyone, particularly guests, who read threads here at MHB. It may look daunting at first, I understand, I felt the same way when I first got involved with math sites, but now I think in $\LaTeX$ and it comes very naturally.

 

[mathdad]

Another issue we need to address here and now is getting you to use $\LaTeX$. What you posted is hard to read for these poor old eyes...(Crying)

$\LaTeX$ needs to be wrapped in tags...the easiest to use are the MATH tags, which can be generated by clicking the $\Sigma$ button on our toolbar. Once you have those tags created, your cursor will be located within them, ready to accept your code. You want the code:

t=\frac{-v\pm\sqrt{v_0^2-4\left(-\frac{1}{2}a\right)\left(-\Delta x\right)}}{2\left(-\frac{1}{2}a\right)}=\frac{v\pm\sqrt{v_0^2-2a\Delta x}}{a}

This, within the MATH tags, produces:

\(\displaystyle t=\frac{-v\pm\sqrt{v_0^2-4\left(-\frac{1}{2}a\right)\left(-\Delta x\right)}}{2\left(-\frac{1}{2}a\right)}=\frac{v\pm\sqrt{v_0^2-2a\Delta x}}{a}\)

Take a bit of time to see how this works. You will find that just about every symbol/command you will ever need is located on our Quick $\LaTeX$ element to help you construct your equations/expressions. (Yes)

Why am I pushing you to learn this? You will find people will be more willing to jump in and help when they can more easily read your work. It just looks better, and is more informative for everyone, particularly guests, who read threads here at MHB. It may look daunting at first, I understand, I felt the same way when I first got involved with math sites, but now I think in $\LaTeX$ and it comes very naturally.

Can the LaTex tool be used via cell phone? I do not have a computer, laptop or tablet. I have a Samsung Galaxy J7 cell phone.

Question:

How does (-1/2)a • 2 = a in the denominator of the quadratic formula created? Shouldn't it be -a?

 

[MarkFL]

Can the LaTex tool be used via cell phone? I do not have a computer, laptop or tablet. I have a Samsung Galaxy J7 cell phone.

I don't use mobile devices, but you should be able to use it. I know others here use mobile devices and $\LaTeX$.

Question:

How does (-1/2)a • 2 = a in the denominator of the quadratic formula created? Shouldn't it be -a?

I factored out -1 from both the numerator and denominator. I also corrected a typo where I had $v$ instead of $v_0$. :)

 

[mathdad]

Ok. Great. Let us move on. I will post similar word problems in the coming days.

 

[MarkFL]

Okay, so we've found:

\(\displaystyle t=\frac{v_0\pm\sqrt{v_0^2-2a\Delta x}}{a}\)

Suppose we have $\Delta x=0$. At what times does that occur?

 

[mathdad]

Okay, so we've found:

\(\displaystyle t=\frac{v_0\pm\sqrt{v_0^2-2a\Delta x}}{a}\)

Suppose we have $\Delta x=0$. At what times does that occur?

Replace (delta)x with 0 and simplify, right?

 

[MarkFL]

Replace (delta)x with 0 and simplify, right?

Yes. :)

 

[mathdad]

Thank you for all your help.