How Far Does a Ball Travel Horizontally Before Hitting the Ceiling?

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Homework Help Overview

The problem involves a ball being thrown at an angle of 45 degrees with a specific initial velocity, and the goal is to determine how far it travels horizontally before hitting the ceiling of a room. The context is rooted in kinematics and projectile motion.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the setup of the problem using the equation for projectile motion and attempt to express the height in terms of the horizontal distance. There are questions about the validity of the quadratic equation derived and concerns regarding the dependence on the variable D.

Discussion Status

The discussion is ongoing, with participants exploring different interpretations of the quadratic equation and its coefficients. Some guidance has been offered regarding the application of the quadratic formula, but there is no consensus on the correct approach or interpretation of the results.

Contextual Notes

Participants are grappling with the implications of the variables involved, particularly the height D, and how it factors into the solution. There is a noted confusion regarding the cancellation of terms in the quadratic formula, which may affect the understanding of the problem.

armolinasf
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Homework Statement


a ball is thrown from the floor at an angle of 45 degrees and with an initial velocity of sqrt(6gD) where D is the height of the room. How far does the ball travel horizontally before hitting the ceiling?

Homework Equations



y=tan(a)x-gx^2/(2v^2cos(a)^2)

The Attempt at a Solution



I let the height equal D

D=tan45x-gx^2/(2sqrt(6gD)^2cos(45)^2)

D=x-(x^2/6D) Solving this as a quadratic equation I (get 3+/-sqrt3)/2. But this answer makes no sense since it should be in terms of D. Where am I going wrong? thanks
 
Last edited:
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armolinasf said:

Homework Statement


a ball is thrown from the floor at an angle of 45 degrees and with an initial velocity of sqrt(6gD) where D is the height of the room. How far does the ball travel horizontally before hitting the ceiling?

Homework Equations



y=tan(a)x-gx^2/(2v^2cos(a)^2)

The Attempt at a Solution



I let the height equal D

D=tan45x-gx^2/(2sqrt(6gD)^2cos(45)^2)

D=x-(x^2/6D) Solving this as a quadratic equation I (get 3+/-sqrt3)/2. But this answer makes no sense since it should be in terms of D. Where am I going wrong? thanks

Show us how you got the answer that doesn't involve D using the quadratic equation.
 
0=-D+tan(45)x-gx^2/(2sqrt(6gD)^2cos(45)^2), tan45=2cos(45)^2=1 and g will cancel. that leaves this quad equation:

-D+x-x^2/(6D)

using the quad formula:

-x +/-(sqrt(1-4(-d)(2/6D))/(4/3)

which simplifies to 3+/-sqrt3)/2
 
armolinasf said:
0=-D+tan(45)x-gx^2/(2sqrt(6gD)^2cos(45)^2), tan45=2cos(45)^2=1 and g will cancel. that leaves this quad equation:

-D+x-x^2/(6D)

using the quad formula:

-x +/-(sqrt(1-4(-d)(2/6D))/(4/3)

which simplifies to 3+/-sqrt3)/2

That's a little chaotic. The quad formula is (-b+/-sqrt(b^2-4ac))/(2a). What are a, b and c in that formula? None of them have an 'x' in it. But the (2a) part has a D in it, doesn't it?
 
a=-(1/6D) b=1 c=-D
 
armolinasf said:
a=-(1/6D) b=1 c=-D

Alright. So what's (-b+/-sqrt(b^2-4ac))/(2a)? I think it has a D in it. The D's cancel in the numerator, not in the denominator.
 
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