Question
A ball is thrown horizontally from the top of a 60.0-m building and lands 100.0 m from the base of the building. Ignore air resistance. (a) How long is the ball in the air? (b) What must have been the initial horizontal component of the velocity? (c) What is the vertical component of the velocity just before the ball hits the ground? (d) What is the velocity (including both the horizontal and vertical components) of the ball just before it hits the ground?

a) $3.50 \textrm{ s}$

b) $28.6 \textrm{ m/s}$

c) $-34.3 \textrm{ m/s}$

d) $44.6 \textrm{ m/s, } 50.2^\circ \textrm{ below the + x-axis}$

# OpenStax College Physics, Chapter 3, Problem 27 (Problems & Exercises)

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Video Transcript

Hello! My version of the first equation you use is $y=y0+v0t-1/2at^2$ , but you always show it with as $y=y0+v0t+1/2a^2$ with a plus sign between $v0t$ and $1/2a^2$ instead of a minus. Why is this?

Hi Helen, good question. I think, but correct me if I'm wrong, what you meant to say was with a $g$, not an $a$, as $y = y_o + v_ot - 1/2 g t^2$. Would you agree? The version I'm showing is more generic, and $y$ can represent any position, not just the vertical position that $y$ is traditionally associated with. In your version, however, $y$ specifically refers to the vertical direction, assumes that upwards is the positive direction (and consequently down is negative), and that the acceleration takes place near the surface of Earth, in which case $a$ becomes $-g$. The short answer is that your equation is the same as mine, except with $-g$ substituted in place of $a$.
All the best,
Shaun

For students who don't mind a little risk and potentially save time: Part C would be easier with Vy=V0y-gt as the equation, however it does rely on data that we gathered through doing the problem, which is probably why Shaun avoids trying to go the easy route.

Shaun, thanks for all these great study materials. Teaching myself Physics online would be significantly harder without your teaching resources.

Does knot stand for initial?

Hi lightspeed, that's a great question. "knot" is, in this case, spelled "naught", which is an older English way of saying "zero". As you correctly surmise, it means "initial" in the context of a physics solution.

to find the Vox velocity we used the x=Xo + Voxt +1/2at^2. Why did you include a when it is o when considering horizontal?

At 7:26 you say there is no acceleration horizontally. Doesn't there have to be horizontal acceleration since someone threw the ball?

Hi tombreaz,
Thank you for the question. You're absolutely right that someone initially accelerated the ball horizontally to make it move sideways. However, the question considers the moment after that. There is no horizontal acceleration once the ball leaves the person's hand, so we can correctly say $a_x = 0$ for the period considered in the solution. The "initial horizontal component of the velocity" found in part (b) is the final horizontal component of the velocity during the ball-in-hand initial acceleration, but our analysis begins after the ball-in-hand throw is done.
Hope this helps,
Shaun

the video won't load for me.

Thank you so much both of you for noticing this video wouldn't load. I had to change its configuration, and it should load fine now.
All the best,
Shaun

Hi! Thank you for the videos! I was just wondering why in part B you would multiply the square root of blah blah blah instead of deciding it away from Vox?

Thank you!

Hello, I might not fully understand your question where you mention "deciding it away from Vox". $\dfrac{\sqrt{-2y_\textrm{o}}}{a_\textrm{y}}$ is the time to fall. That's $t$ in other words. Dividing the horizontal distance covered by that time tells us the initial (and constant) horizontal component of the velocity. In the video near 2:52 you'll see $x$ multiplied by the reciprocal of $t$, which is the same as dividing by $t$. Does that help?
All the best,
Shaun

Hello, I am trying to learn these equations and I wanted to know where is the -2 coming from in problem A and B that you are dividing?

Hi timmany, thank you for the question. The negative sign comes from moving $y_o$ to the other side of the equation - or put in other words, by subtracting $y_o$ from both sides in which case it becomes $-y_o$ on the side where it didn't originally exist. The 2 results from multiplying both sides by 2 as part of the algebra to isolate $t$; this is explained at 1:41 in the video.
All the best,
Shaun

Hi. in part B of the problem when you were solving for Vo(x) I noticed that right before you plugged in the respective numbers, it went to the inverse. It went from the square root of -2y(o)/a(y) to the square root of a(y)/-2y(o). Why is that?

Thank you for the question k.miller. At 3:44 I make the algebra step of solving for $v_{\textrm{ox}}$. $v_\textrm{ox}$ is being multiplied by something, so to isolate it you need to divide both sides by that factor. The factor is $\sqrt{\dfrac{-2y_\textrm{o}}{a_\textrm{y}}}$ and I show dividing by it as multiplying by its reciprocal. Dividing by something is the same as multiplying by its reciprocal and as a matter of personal preference I don't like dividing fractions by fractions since I find that looks messy. The factor $\sqrt{\dfrac{a_\textrm{y}}{-2y_\textrm{o}}}$ is the reciprocal, and I'm multiplying both sides by it, and it cancels with the factor $\sqrt{\dfrac{-2y_\textrm{o}}{a_\textrm{y}}}$ on the side with $v_\textrm{ox}$ thereby isolating and solving for it. Hope this helps!
Shaun