# Homework Help: Projectile motion Question

1. Jul 1, 2011

### armolinasf

1. The problem statement, all variables and given/known data
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?

2. Relevant equations

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

3. 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: Jul 1, 2011
2. Jul 1, 2011

### Dick

Show us how you got the answer that doesn't involve D using the quadratic equation.

3. Jul 1, 2011

### armolinasf

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)

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

which simplifies to 3+/-sqrt3)/2

4. Jul 1, 2011

### Dick

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. Jul 1, 2011

### armolinasf

a=-(1/6D) b=1 c=-D

6. Jul 1, 2011

### Dick

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.

Last edited: Jul 1, 2011