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

In summary: Using the quadratic formula, the answer should be (-b+/-sqrt(b^2-4ac))/(2a), which simplifies to (1+/-sqrt(1+4D))/(1/6D), which can be rewritten as 6D+/-sqrt(36D^2+4D)/D, which can be further simplified to (6+/-sqrt(36+4/D))/D. So the answer is (6+/-sqrt(36+4/D))/D, not (3+/-sqrt(3))/2.
  • #1
armolinasf
196
0

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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  • #2
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.
 
  • #3
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
 
  • #4
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?
 
  • #5
a=-(1/6D) b=1 c=-D
 
  • #6
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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Related to How Far Does a Ball Travel Horizontally Before Hitting the Ceiling?

1. What is projectile motion?

Projectile motion is the motion of an object that is thrown or launched into the air and moves under the influence of gravity.

2. What are the factors that affect projectile motion?

The factors that affect projectile motion include the initial velocity, angle of launch, air resistance, and gravity.

3. How do you calculate the range of a projectile?

The range of a projectile can be calculated using the equation: R = (v^2 * sin(2θ)) / g, where R is the range, v is the initial velocity, θ is the angle of launch, and g is the acceleration due to gravity.

4. What is the maximum height of a projectile?

The maximum height of a projectile can be calculated using the equation: h = (v^2 * sin^2(θ)) / 2g, where h is the maximum height, v is the initial velocity, θ is the angle of launch, and g is the acceleration due to gravity.

5. How does air resistance affect projectile motion?

Air resistance can affect projectile motion by slowing down the object's motion and changing its trajectory. This is because air resistance is a force that acts opposite to the direction of motion, causing the object to lose speed and change direction.

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