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?

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Final Answer

a) $3.50 \textrm{ s}$

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

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

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

Solution video

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

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This is College Physics Answers with Shaun Dychko. This ball is thrown horizontally from the top of a building. The building we'll say is 60 meters high, so we'll say that y naught equals 60. The initial y position is 60 and we'll take the ground to be y equals zero. There is no acceleration horizontally but there is gravity and that's going to affect the vertical direction and so the vertical acceleration is negative 9.8 meters per second squared. Initially there is no vertical velocity at all because it says the ball is thrown horizontally. So there is some v subscript x naught but we don't know what that is. That's one of the questions is to find it. So part A asks us for how long is the ball in the air. So we can use this formula to figure that out, the final y position is the initial y position plus the initial y velocity times time, plus one half times the vertical acceleration times time squared. This term is zero because there is no initial vertical velocity and this term here is also zero because the final position is zero. So we can subtract y naught from both sides, then switch the sides around and we get one half ayt squared equals negative y naught. Then, our job here by the way is to solve for t because the question is asking for how long is the ball in the air. So we will isolate t by multiplying both sides by 2 over ay and then we end up with this line t squared equals negative two times the initial y position, divided by the vertical acceleration. Then take the square root of both sides and mathematically speaking we should put a plus or minus here, but physically speaking, we know that the time needs to be positive, there's no physical interpretation for a negative time here. So we'll just take the positive square root. So time is the positive square root of negative two y naught over ay so that's the square root of negative two times 60 divided by negative 9.8 meters per second squared, giving us 3.50 seconds in the air. The next question is asking how far from the base of the building will the ball land. So what is x in other words or, wait a second, that's not the question. No, v naught, yeah, we're given x actually. It's 100 meters. We're told that the ball lands 100 meters from the base of the building. So, v naught x is the question here. Well, we can use this formula because the only unknown is v naught x in here because we know t and we figured it out before in part A and we know what x naught is, it's zero because it starts at position zero. So this is v naught x times the square root of negative two y naught over ay. I could have just written the number 3.5 here but I'm just trying not to use the result of previous parts of the question in subsequent parts and trying to just come up with formulas that use the data that we've been originally given in the question. So I wrote square root negative two y naught over ay as the substitution for t. Then we will solve for v naught x by multiplying both sides by ay over negative two y naught square rooted. Then the same on the other side, and then we end up with this line here after you switch the sides around. You have the horizontal initial velocity which you know is the horizontal velocity in general because it's never changing since there's no horizontal acceleration at all. It equals the square root of ay over negative two y naught, all times the final x position x. So that's square root of negative 9.8 meters per second squared over negative two times 60 meters, all times 100 meters, giving us 28.6 meters per second as the initial horizontal velocity. Then part C is asking for the final y component of its velocity and this formula here will be helpful because we know the initial y component of its velocity is zero because the ball is thrown horizontally so there's no component in the y direction. We know the initial y position is 60 meters. We know the final y position is zero and so this is the formula containing only one unknown which is the thing we want to find. So we'll get rid of this term and then take the square root of both sides and we get vy is negative. Then we put the negative there because we know the ball is going downwards and so the y component of its velocity is negative, times the square root of two times vertical acceleration times the final position minus the initial position. So that's negative two times -- sorry, negative square root of two times negative 9.8, times zero minus 60. That gives negative 34.3 meters per second. Now, putting the x and y components together, we can figure out the final velocity just as it's hitting the ground. So the magnitude of its final velocity will be the square root of the sum of the squares of the components of the velocity. So that's square root of negative 34.29 meters per second squared, and notice I've written more digits here in this working than I had in my final answer for the y component of velocity. That's a good best practice because normally you want to avoid rounding numbers that you use in your calculations. If we put in negative 34.3 here, we will get a slightly different answer actually. Probably we get I think 44.7 and that 44.7 would be called an intermediate rounding error because using a rounded number in the calculation introduces its own error. So we will use un-rounded numbers as much as possible. Then plus the horizontal component of its velocity 28.5774 meters per second squared and square rooting that sum of the squares gives 44.6 meters per second. The direction is going to be the inverse tangent of the y component of velocity divided by the x component of the velocity and we're ignoring the negative signs. So that's why I have these absolute value symbol here, these vertical bars. So that is just to say that you take the inverse tangent of 34.2929 divided by 28.5774 giving us 50.2 degrees. Because the velocity at the end here is down to the right and we've taken the inverse tangent of its y component divided by its x component. The angle that we found is this one. So this angle is below the positive x axis and so our final answer for velocity just before the ball hits the ground is 44.6 meters per second, 50.2 degrees below the positive x axis.

Comments

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.

Thanks for helping out your fellow students, and thanks also for the feedback. I'm really glad the solutions are helping!

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?

Hello, at 2:31 I'm seeing the formula without $a$ . Am I missing something?

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.

Same here

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

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

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Chapter 3: Two-Dimensional Kinematics

Problems & Exercises