Throwing a ball from the roof of a building

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Homework Help Overview

The problem involves a ball being thrown horizontally from the roof of a 54 m tall building and landing 50 m away from the base. Participants are exploring the initial speed of the ball using principles of projectile motion.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to identify relevant equations of motion and expresses uncertainty about how to proceed with the problem. Some participants suggest using the equations of motion to relate distance, velocity, and time, while others provide a method to calculate the time of fall and subsequently the horizontal velocity.

Discussion Status

Participants have shared various approaches to the problem, including deriving time from the vertical motion and using it to find horizontal velocity. The original poster has performed calculations based on the suggested methods and is seeking validation of their results. There is no explicit consensus on the correctness of the calculations, but guidance has been provided on how to check the answer.

Contextual Notes

The original poster is constrained by a single submission for the question and expresses a desire to understand the process for similar future problems.

lando45
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A ball is thrown horizontally from the roof of a building 54 m tall and lands 50 m from the base. What was the ball's initial speed?

I have been set this question and I'm a little stuck...here is the information I have extracted from the question:

sv = -54
sh = 50
v = ?
a = -9.81

I don't know what else I can do, or what formulas/principles I should be using. I can't modulate with Pythagoras as surely the ball wouldn't travel downwards in a straight line to it's resting point...any help?

Much appreciated,
Rory
 
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You're familiar with the basic equation of motion relating distance to velocity and time, correct?
[tex]d = d_o + V_o*t+\frac{1}{2}a*t^2[/tex]
where
d_o = original distance
V_o = original velocity
a = acceleration (in our case, gravity)
t = time

You have two unknown variables: Original Velocity and Time
Well, here's what you'd do.
V_o_x = V_o
V_o_y = 0

Because the ball has no initial vertical velocity component, you can solve for the free fall equation [tex]d_f_a_l_l = 0.5g*t^2[/tex]
solving symbolicaly, we obtain [tex]t = \sqrt{\frac{2*d_f_a_l_l}{g}}[/tex]
** Substitute your building height and g (9.81 m/s^2) into this equation and solve for time

Now you can use the second motion equation Distance = Velocity * Time to achieve the original horizontal velocity.

Look at it this way; the ball travels a known distance horizontally, and it does it in a known amount of time, because we know how long it would take for the ball to fall. Because there is no horizontal acceleration, we can solve the horizontal velocity equation relatively easily simply by setting:
[tex]V_o = d_h / t_f_a_l_l[/tex]
which gives us
[tex]V_o = \frac{d_h}{\sqrt{\frac{2*d_f_a_l_l}{g}}}[/tex]

I know it looks nasty, but it's really quite easy once you start getting some solid numeric values. I'll leave the calculations up to you.
 
Wow thanks for that. Ok so this is what I've done now:

t = sqrt [ (2 * d_fallen) / g ]
t = sqrt [ (2 * 54) / 9.81 ]
t = sqrt [ 108 / 9.81]
t = sqrt 11.009
t = 3.318

d_horizontal = vt
v = s / t
v = 50 / 3.318
v = 15.069

Does this seem like the correct answer? Cos I only have one submission for this question, and I really don't want to get it wrong, I want to know how to answer questions like this! So does 15.069 ms^-1 seem about right? Or have I gone wrong somewhere in my calculations...
 
Approximation, estimation, generalization

if you're unsure about your answer, you can do one of two things to check it...
A) Solve it backwards using your answer.
B) Graph it. It's easy, really... just set X as the time variable and d_y as the y variable, and you should easily be able to see if your answer for time @ y=0 is correct...
attachment.php?attachmentid=5735&stc=1&d=1133478133.png

Learn to love your TI-83, 83+, 86, 89... :approve:
 

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ah thanks a lot i got the answer correct, much appreciated.
 

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