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

Solution Video

# OpenStax College Physics Solution, Chapter 3, Problem 27 (Problems & Exercises) (7:49)

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21 votes with an average rating of 3.5.

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Video Transcript

Submitted by Helen on Tue, 02/11/2020 - 08:15

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?

Submitted by ShaunDychko on Tue, 02/11/2020 - 09:50

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

Submitted by georgeh on Sun, 04/26/2020 - 10:06

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.

Submitted by lightspeed on Fri, 09/04/2020 - 20:37

Does knot stand for initial?

Submitted by ShaunDychko on Mon, 09/07/2020 - 22:22

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.

Submitted by danhthang193 on Thu, 09/10/2020 - 16:14

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?

Submitted by tombreaz on Thu, 10/01/2020 - 20:36

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

Submitted by ShaunDychko on Wed, 10/21/2020 - 08:02

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

Submitted by FLS1036 on Tue, 01/26/2021 - 15:54

the video won't load for me.

Submitted by ShaunDychko on Wed, 01/27/2021 - 10:14

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