Throwing a ball from the roof of a building

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A ball thrown horizontally from a 54 m tall building lands 50 m from the base, prompting a calculation of its initial speed. The relevant equations of motion were discussed, focusing on the relationship between distance, velocity, and time under gravity. By calculating the time it takes for the ball to fall using the formula t = sqrt(2*d/g), the time was determined to be approximately 3.318 seconds. Subsequently, the horizontal speed was calculated as v = d/t, yielding an initial speed of about 15.069 m/s. The calculations were confirmed to be correct, demonstrating a successful application of the principles of projectile motion.
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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?
d = d_o + V_o*t+\frac{1}{2}a*t^2
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 d_f_a_l_l = 0.5g*t^2
solving symbolicaly, we obtain t = \sqrt{\frac{2*d_f_a_l_l}{g}}
** 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:
V_o = d_h / t_f_a_l_l
which gives us
V_o = \frac{d_h}{\sqrt{\frac{2*d_f_a_l_l}{g}}}

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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