Throwing a ball from the roof of a building

In summary, the ball had an initial speed of -9.81 ms^-1 when it was thrown from the roof of the building 54 m tall.
  • #1
lando45
84
0
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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  • #2
(forenote, an underscore indicates a subscript)

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.
 
  • #3
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...
 
  • #4
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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  • #5
ah thanks a lot i got the answer correct, much appreciated.
 

1. How far will a ball travel when thrown from the roof of a building?

The distance a ball will travel when thrown from the roof of a building depends on several factors, such as the initial velocity, angle of release, and air resistance. Using the equation d = v0t + 1/2at2, where d is distance, v0 is initial velocity, a is acceleration due to gravity (9.8 m/s2), and t is time, the ball will travel further the faster it is thrown and the lower the angle of release.

2. Will a heavier ball travel further when thrown from the roof of a building?

Assuming all other factors remain constant, a heavier ball will not necessarily travel further when thrown from the roof of a building. The distance a ball travels is determined by its initial velocity and angle of release, not its weight. However, a heavier ball may be able to withstand air resistance better and maintain its speed for longer, resulting in a slightly longer distance.

3. How does air resistance affect the distance a ball travels when thrown from the roof of a building?

Air resistance, also known as drag, acts in the opposite direction of motion and can decrease the distance a ball travels when thrown from the roof of a building. As the ball moves through the air, it experiences a force that slows it down. This effect is more significant for lighter and less aerodynamic objects, such as a feather, compared to a dense and streamlined object, like a baseball.

4. Is it possible for a ball thrown from the roof of a building to reach escape velocity?

No, it is not possible for a ball thrown from the roof of a building to reach escape velocity. Escape velocity is the speed required for an object to overcome the gravitational pull of a celestial body, such as Earth. The escape velocity of Earth is approximately 11.2 km/s, which is much faster than any human can throw a ball.

5. Can the height of the building affect the distance a ball travels when thrown from the roof?

Yes, the height of the building can affect the distance a ball travels when thrown from the roof. The higher the building, the longer the ball has to accelerate due to gravity, resulting in a greater distance traveled. However, this effect is relatively small compared to other factors, such as initial velocity and angle of release.

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