# Applying kinematics issue

I've been having an unusually difficult time solving kinematics problems in comparison to all the other students in the classroom. It appears that I'm one of the unfortunate people with an anti-math mind (although I've been expecting as much for some time now). In short, I have no idea how to solve this problem:

A carpenter tosses a shingle off a $9.4m$ high roof, giving it an initial horizontal velocity of $7.2 m/s$.
a. What is the final vertical velocity of the shingle?
b. How long does it take to reach the ground?
c. How far does it move horizontally in this time?

I tried to organize the data I had and the data I needed to calculate:

Horizontal:
1. time is unknown and the same as vertical time
2. initial position is $0m$
3. final position is unknown
4. initial velocity is $7.2m/s$
5. final velocity is $7.2m/s$
6. acceleration must be $0$ because velocity doesn't change

Vertical:
1. time is unknown and the same as vertical time
2. initial position is $12m$
3. final position is $0m$
4. initial velocity is gravity, or $9.81m/s$
5. final velocity is unknown
6. acceleration is $9.81m/s^2$

However, according to my unknowns and my equations, I cannot ultimately solve any of these because I cannot solve for time, which requires knowing distance, velocity, and acceleration, like this:
$t = -v_i \pm \sqrt{v_i^2 + 2a(\pm d)} \over{a}$ Is this correct and is there another way to find time with less unknowns?

And none of my equations can explicitly find the initial or final positions of anything:

$\Delta x = v_{ix} \Delta t + 0.5a_x \Delta t^2$ (also works with $y$)
$v_{fx}^2 - v_{ix}^2 = 2a_x \Delta x$ (also works with $y$)
$v_{fx} = v_{ix} + a_x \Delta t$ (also works with $y$)
$\Delta y = 0.5(v_{fy} + v_{iy}) \Delta t$
$y = v_{avg} * t$
$\Delta y = 0.5(V_{fy} + V_{iy}) \Delta t$

What I want to know is how to correctly solve this (with equations listed). Please do not just provide the answer; thanks in advance.

• I wrote up a long answer to this and then realized you basically didn't know you were supposed to assume the initial vertical velocity is $0$. Oh well. I think my answer is ok anyway – Brian Moths Nov 6 '13 at 6:16
• All five equations work for horizontal and vertical motion; just make horizontal acceleration zero. – DJohnM Nov 6 '13 at 8:15

Let's solve the problem:

# Step 1: What is going on?

We are standing on a roof and we throw an object of horizontal. The object will fly through the air and will eventually hit the ground because of gravity. It would be good to draw a picture at this point but I am too lazy.

# Step 2: What do we want to know?

We want to know the final vertical velocity of the shingle, how long until the shingle hits the ground, and how far the shingle moves vertically (how far the shingle lands from the base of the building).

# Step 3: What are we given?

We are told the height of the building and how fast we throw the shingle.

# Step 4: How do we expect what we want to know to depend on what we know?

Well if the building gets shorter, I would expect the time to fall to be less, and I would expect the shingle to hit the ground with less vertical velocity. I also expect the shingle to not land as far away from the building because it won't fly for as long.

For example consider the extreme case where the height the shingle falls is zero. Then the shingle hits the ground immediately so the fall time is zero and its velocity is zero when it hits the ground. Also the shingle doesn't move horizontally at all.

Now what if we throw it harder. I am not exactly sure how this will affect fall time or the velocity it hits the ground. Maybe it will stay in the air longer because you are throwing it so hard. Maybe it will hit the ground faster because it is moving faster. Similarly I am not sure how the vertical impact speed will be affected. But I am pretty sure that the harder I throw it, the farther it will land from the base of the building. For example, if I throw it with zero speed, it will just go straight down and there will be no horizontal displacement.

# Step 5: What physical concepts are at play to determine the relationships described in the previous step?

Well we said that the taller the building is the longer the time the shingle will fall. The concept behind this must have something to do with gravity, because gravity is what is making the shingle fall. Gravity is also responsible for the shingle having a final velocity.

Now what about the horizontal motion? This must have something to do with the initial horizontal speed. So there must be some law relating a speed and a distance traveled.

# Step 6: What are the relevant laws having to do with these concepts?

We said the shingle was going to fall because of gravity. We would like a quantitative law explaining how exactly the shingle falls because of gravity. The best we could ask for is having vertical displacement as a function of time. Fortunately this law is known. It goes like this $$\Delta y(\Delta t) = v_{0y}\Delta t - \frac{1}{2} g (\Delta t)^2,$$ where $\Delta y(t)$ is the $y$ displacement after a time $\Delta t$, $v_{0y}$ is the initial velocity in the $y$ direction, and $g$ is the acceleration of gravity and we have taken up to be in the positive $y$ direction. Notice that the displacement in the $y$ direction doesn't care about velocity in the $x$ direction.

We can now ask what we know and don't know for our problem. We do know $\Delta y$ (the height of the building), $v_{0y}$ (it is zero), and $g$. We don't know $\Delta t$. Fortunately this is the only unknown in the equation. Thus we will be able to solve for $\Delta t$ and we will then know how long it takes the object to fall.

Another thing we need to find is the final velocity of the shingle. The velocity was also due to gravity, so we must find a law relating a velocity to gravity. We know that gravity causes constant acceleration, so the change in is proportional to time. The constant of proportionality is the acceleration. Thus we get the law $$\Delta v_y(\Delta t) = -g \Delta t.$$ At this point, the only unknown is $\Delta v_y$, so this equation determines that value.

The other thing we need to find is the horizontal displacement of the ball. We know the initial speed we threw it at, and we need its horizontal displacement. There is nothing causing the shingle to accelerate besides gravity, and we know gravity only affects motion in the vertical direction and not the horizontal direction, so the law we want to use is $$\Delta x (\Delta t) = v_{0x} \Delta t,$$ where $\Delta x$ is the change in the $x$ position and $v_{0x}$ is the initial velocity in the $x$ direction. The only unknown in this equation is $\Delta x$, so this equation determines $\Delta x$.

# Step 7: Use the laws to find the answers

Our first law is $\Delta y(\Delta t) = v_{0y}\Delta t - \frac{1}{2} g (\Delta t)^2$. For us, $\Delta t$ is $t_f$, the time it takes the shingle to fall, and $\Delta y(\Delta t)$ is $-H$, where $H$ is the height of the building. We are told that the shingle is thrown horizontally so $v_{0y}$ is zero. The law then becomes $-H = - \frac{1}{2} g t_f^2.$ We were asked for $t_f$. Solving for this we find $$t_f = \sqrt{\frac{2 H}{g}}.$$

Next we will use our second law to find the final $y$ velocity of the shingle. The second law was $\Delta v_y(\Delta t) = -g \Delta t.$ Again, $\Delta t$ is $t_f$. Our $\Delta v_y(\Delta t)$ is the change actually just the final $y$ velocity $v_{fy}$ since the initial $y$ velocity is $0$. Thus we find $$v_{fy} = -g t_f = -g\sqrt{\frac{2 H}{g}} = -\sqrt{2 H g}.$$

Last we should use the law for the horizontal displacement. This law was $\Delta x (\Delta t) = v_{0x} \Delta t.$ In our case $\Delta x (\Delta t)$ is the horizontal displacement of the shingle when it hits the ground $x_f$. $v_{0x}$ is the horizontal velocity we threw the shingle at. The law we get is $$x_f = v_{0x} t_f = v_{0x} \sqrt{\frac{2 H}{g}}.$$

# Step 8: Check dimensions of our answers

We check our answers. We will first check the dimensions. We will use $L$ to denoted a dimension of length and $T$ to denote a dimension of time.

We will start with $v_f$. We found $t_f = \sqrt{\frac{2 H}{g}}$. The dimension we get for $t_f$ are $\sqrt{\frac{ L}{L/T^2}} = \sqrt{T^2} = T$, so we get the right dimensions.

Now let's check dimensions on $v_{fy}$. We found $v_{fy} = -\sqrt{2 H g}$. Thus the dimensions of $v_{fy}$ are $\sqrt{L L/T^2} = \sqrt{(L/T)^2} = L/T$, which is the dimensions for speed.

Finally let's check the dimensions on $x_f$. We found $x_f = v_{0x} \sqrt{\frac{2 H}{g}}$. The units for $x_f$ are $L/T \sqrt{\frac{L}{L/T^2}} = L/T \sqrt{T^2} = L/T * T = L$. These units make sense.

# Step 9: Check answer scales as expected.

We should check to make sure the things we are being asked for depend on what we are given as we expected in Step 4.

Let's start with $t_f = \sqrt{\frac{2 H}{g}}$. We see that $t_f$ gets bigger when $H$ gets bigger as we expected. We also find that it has no dependence on $v_{0x}$ which is what we realized should be the case. Also we see that if $g$ were to get stronger $t_f$ would decrease, which makes sense.

Now consider $v_{fy} = -\sqrt{2 H g}$. We see that as $H$ gets bigger $v_{fy}$ gets bigger as expected. Also we see that if gravity were to get stronger, $v_{fy}$ would also increase. This makes sense.

Finally consider $x_f = v_{0x} \sqrt{\frac{2 H}{g}}$. We see that if we were to throw the shingle harder it would go farther, as expected. Also the higher the building is (bigger $H$), the farther the shingle goes, as expected. If gravity were to get stronger, we predict the shingle wouldn't go as far. This makes sense because the shingle would hit the ground a lot sooner.

# Step 10: plug in numbers

Now is a good time to plug in numbers (Note steps 8 and 9 are impossible if you plug in numbers too soon). We find $t_f = 1.4 \mathrm{\ s}$, $v_{fy} = -14 \mathrm{\ m/s}$, and $x_f = 10. \mathrm{\ m}$.

I think an important thing about solving the problems is doing the first five steps. Usually these tell you a lot about what equations you should think about using. Another important this is to not bother plugging in numbers until the very end.