Calculating the horizontal distance traveled of an object that was thrown.

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Homework Help Overview

The problem involves calculating the horizontal distance traveled by a ball thrown onto a roof, given the time of flight and the height at which it lands. The subject area pertains to kinematics and projectile motion.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to calculate the horizontal distance using kinematic equations and trigonometry, but expresses confusion over the results. Some participants question the clarity of the problem setup and the initial conditions, while others suggest focusing on vertical motion first.

Discussion Status

The discussion is ongoing, with participants providing guidance on how to approach the problem without giving complete solutions. There is an emphasis on ensuring that the original poster understands the vertical motion before integrating horizontal components.

Contextual Notes

There is a request for clarification regarding the initial conditions of the ball's motion, including the angle of projection and the nature of the throw. The original poster's calculations are noted to be inconsistent with expected results, prompting further inquiry.

DavidAp
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A ball is thrown up onto a roof, landing 3.70s later at height h = 21.0m above the release level. The ball's path just before landing is angled at θ = 54.0˚ with the roof.
Find the horizontal distance it travels.

Answer: 33.5m


Related Equations:
Xf = Xo + 1/2(Vf + Vo)t
_____________________

My approach on this problem was to find the velocity the moment before impact with the roof and to use basic trigonometry to calculate the horizontal velocity of the ball. From the given facts I can infer that:

Xf = 21m
Xo = 0m
t = 3.70s
Vo = ?
Vf = Vo*sin(54)

Now, plugging this into the equation above I get:
21 = 1/2(Vo*sin(54) + Vo)3.70
42 = (Vo*sin(54) + Vo)3.70
11.35 = .81Vo + Vo
11.35 = 1.81Vo
6.27 = Vo

With Vo I can calculate the horizontal velocity which would be:
Vx = Vo*cos(54)
Vx = 6.27(.59)
Vx = 3.69

And the distance should therefore be:
d = VxT
d = 3.69(3.70)
d = 13.65

However, the answer is not 13.65m and after scratching my head and approaching this problem through multiple, yet similar, angles I finally decided to ask for help. Can somebody help me and point out what I did wrong? I thought I had the right idea but I can't seem to get the right answer?

Thank you for taking the time to read my question.
Sincerely David A.P.
 
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dave, can you write the original question AS IT IS. its not clear how the ball is flying. was it thrown vertically upwards or what ? also when it lands on the roof, which angle the \theta is ? a pic will help
 
Sure! Here's an image that came with the problem. However, the wording is nearly identical to the problem itself.

14ahqba.jpg
 
Concentrate on the vertical motion first. Rather than clutter things with the sine and cosine functions from the outset, just assume that the initial vertical velocity is vyo. Can you solve for it using one of the common kinematic equations from what you're given?
 
IssacNewton said:
hi david

here's how i would do it.
[SNIP ALL]

IssacNewton, it is a Physics Forums rule that helpers NOT simply solve homework problems and present complete solutions. Posters must do their own work, with help, hints, and guidance provided as required. Breaking the rule can get one banned from PF.
 

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