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

AI Thread Summary
A ball thrown at a 45-degree angle with an initial velocity of sqrt(6gD) will travel horizontally until it hits the ceiling at height D. The equation used to determine the horizontal distance is y=tan(a)x-gx^2/(2v^2cos(a)^2), which simplifies to a quadratic equation. The attempt to solve this resulted in a confusing answer that did not include D, leading to questions about the quadratic formula's application. Clarification revealed that the coefficients in the quadratic formula do indeed involve D, and the D's cancel in the numerator but not in the denominator. The discussion highlights the importance of correctly identifying and applying the quadratic formula in physics problems involving projectile motion.
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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
 
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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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