How Do You Calculate the Initial Velocity and Tower Height in Projectile Motion?

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Homework Help Overview

The discussion revolves around a projectile motion problem involving a rock thrown from the Leaning Tower of Pisa at an angle. Participants are tasked with calculating the initial velocity and the height of the tower based on given parameters, including flight time and horizontal distance traveled.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the use of the range formula and question its applicability due to differing initial and final heights. There are attempts to calculate the x and y components of velocity using trigonometric relationships and basic kinematics. Some participants express confusion about the height of the tower and the validity of their calculations.

Discussion Status

Several participants have provided insights into the calculations and have raised questions about the assumptions made in the problem. There is acknowledgment of differing results, with some participants suggesting that the provided height of the tower may not be accurate. The conversation remains open-ended with no clear consensus on the correct approach or answer.

Contextual Notes

Participants note discrepancies between their calculated heights and the commonly accepted height of the Leaning Tower of Pisa, leading to discussions about the accuracy of the problem's parameters. There is also mention of potential constraints regarding the use of calculus in solving the problem.

scorpion990
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Galileo throws a rock from the top of the Leaning Tower of Pisa at an upward angle of 60 degrees with speed v. The rock is in flight for 6.5 seconds and hits the ground 15 m from the base of the building. Ignore air resistance and ignore the fact that the tower tilts a bit. a. What is the speed v? b. How high off the ground is the top of the tower?

This is what I did the first time:
R = v^2*sin(theta)/g (Formula for range)
15 = v^2*sin(60)/9.8
v is approximately 13.028468 m/s
vy = 13.028468sin60 = 11.2829843 m/s
I then used the fact that ay = -9.8 m/s^2 and integration to make a model for position. I set it equal to 0, and I got that the height of the building equals 133 m. However, the real height is about 50 m, so I know I'm doing something incorrectly =/

I just realized that my logic is flawed. The "range" is defined as the distance in the x direction traveled where the starting and ending heights are the same. Mine are not, so I can't use the range formula. However, I can't imagine how else I can solve this. Can anybody point me in the right direction? Thanks.
 
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The horizontal speed of the rock is constant and it is the x-component of the initial speed, v, of the rock.
 
Start by computing the x-component of the speed. Like andrevdh says, its consistent. So it's no different than asking "how fast do I need to go to cover 15 meters in 6.5 seconds?" The easy way to think of this is the speedometer on your car. If you are traveling 60 miles per hour, that's 60/1, which is your distance divided by your time. So velocity is distance / time. 60 miles in 2 hours = 60/2 or 30 mph.

Once you know your velocity's x-component (15m/6.5s), and your angle (60 deg), draw a triangle, use some trig, and compute the other leg of the triangle. It represents your velocity's y-component.

Also avoid doing stuff like this: 15 = v^2*sin(60)/9.8
Don't start plugging numbers into your formulas until you have isolated your unknown variable to the left of the equal sign. Assuming this is the formula you would want to use:

R=v^2*sin(theta)/g
v^2=R/(sin(theta)/g)
v=sqrt(R/(sin(theta)/g))

Now plug in your numbers:

v=sqrt(15/sin(60)/9.8))

It's not wrong to do it your way, but it leads to more errors because you're manipulating numbers instead of variables.

I forget what the range formula is, but are you sure that's it? your x-component would be cos(60) or 1/2 of this speed, and something traveling 6.5 m/s for 6.5 seconds is going to go a lot farther than 15 meters.

scorpion990 said:
I then used the fact that ay = -9.8 m/s^2 and integration to make a model for position.

Does your teacher really want you to use calculus to solve this problem? It's not necessary.

I think your teacher is fudging the numbers here. The real Tower of Pisa is 55 meters tall. Like you, I get an answer a lot higher than that (but not equal to your answer).
 
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Being an old Fire Control Technician, the speed of gravity is 32 feet per second per second in a vacuum at sea level. Just do the math.
 
capnahab said:
Being an old Fire Control Technician, the speed of gravity is 32 feet per second per second in a vacuum at sea level. Just do the math.

That would be the acceleration of gravity, not the speed.


EDIT -- Oh, and the acceleration of gravity does not depend on whether there is a vacuum or not.
 
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I still can't seem to solve it =/
I get
Vx = 2.31 m/s
Vy = 4.00 m/s

Py = -4.9t^2 + 4t + H
Plugging in Py = 0 and t = 6.5 yields a much higher H than 50 m =/ Errrrrrr...
 
It looks like you did it right, is 50m a given answer or for example just a fact
 
Last edited:
I got a much higher answer too. What answer did you get? What makes you think the answer is 50? Is it in the back of the book, or did you Google for Leaning Tower of Pisa and get the answer there?
 
I got ABOUT 180 meters. I don't remember the exact answer. I googled it, and it said H = 50 meters =/

I think I'm just being overly paranoid about getting it correct.
 
  • #10
I got 181 meters. Don't trust that the author of the book knows how tall the Tower of Pisa is.
 

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