I think the real problem is calculating how many beers it would take before someone would actually do this.
x=beers
x(distance of slide)2 + 2x(slope of ramp) + x2(distance to pool) + a six pack = likelihood I would actually do it.
Ric
For us engineers - a bit of fun
Distance travelled
The total horizontal distance (d) traveled.
d = \frac{v \cos \theta}{g} \left( v \sin \theta + \sqrt{(v \sin \theta)^2 + 2gy_0} \right)
When the surface the object is launched from and is flying over is flat, the distance traveled is:
d = \frac{v^2 \sin(2 \theta)}{g}
As a special case, the distance is given by
d = \frac{v^2}{g}
when the angle (θ) is 45° and the initial height (y0) is 0.
For explicit derivations of these results, see Range of a projectile.
[edit] Time of flight
The time of flight (t) is the time it takes for the projectile to finish its trajectory.
t = \frac{d}{v \cos\theta} = \frac{v \sin \theta + \sqrt{(v \sin \theta)^2 + 2gy_0}}{g}
As above, this expression can be reduced to
t = \frac{\sqrt{2} \cdot v}{g}
if θ is 45° and y0 is 0.
The above results are found in Range of a projectile.
[edit] Angle of reach
The "angle of reach" (not quite a scientific term) is the angle (θ) at which a projectile must be launched in order to go a distance d, given the initial velocity v.
\sin(2\theta) = \frac{gd}{v^2}
\theta = \frac{1}{2} \arcsin \left( \frac{gd}{v^2} \right)
[edit] Conditions at an arbitrary distance x
[edit] Height at x
The height y of the projectile at distance x is given by
y = y_0 + x \tan \theta - \frac {gx^2}{2(v\cos\theta)^2} .
The third term is the deviation from traveling in a straight line.
[edit] Velocity at x
The magnitude, | v | , of the velocity of the projectile at distance x is given by
| v | = \sqrt{v^2 - 2gx \tan \theta + \left(\frac{gx}{v\cos \theta}\right)^2} .
[edit] Derivation
The magnitude |v| of the velocity is given by
| v | = \sqrt{V_x^2 + V_y^2} ,
where Vx and Vy are the instantaneous velocities in the x- and y-directions, respectively.
Here the x-velocity remains constant; it is always equal to v cos θ.
The y-velocity can be found using the formula
vf = vi + at
by setting vi = v sin θ, a = g, and t = \frac{x}{v \cos \theta}. (The latter is found by taking x = (v cos θ) t and solving for t.) Then,
V_y = v \sin \theta - \frac{gx}{v \cos \theta}
and
| v | = \sqrt{(v \cos \theta)^2 + \left(v \sin \theta - \frac{gx}{v \cos \theta} \right)^2} .
The formula above is found by simplifying.
[edit] Angle θ required to hit coordinate (x,y)
Vacuum trajectory of a projectile for different launch angles
To hit a target at range x and altitude y when fired from (0,0) and with initial velocity v the required angle(s) of launch θ are:
\theta = \tan^{-1}{\left(\frac{v^2\pm\sqrt{v^4-g(gx^2+2yv^2)}}{gx}\right)}
Each root of the equation corresponds to the two possible launch angles so long as both roots aren't imaginary, in which case the initial velocity is not great enough to reach the point (x,y) you have selected. The greatest feature of this formula is that it allows you to find the angle of launch needed without the restriction of y = 0.
Derivation
First, two elementary formulae are called upon relating to projectile motion:
x = v t \cos \theta , t = \frac{x}{v \cos \theta} (1)
y = vt \sin \theta - \frac{1}{2} g t^2 (2)
Solving (1) for t and substituting this expression in (2) gives:
y = x \tan \theta - \frac{gx^2}{2v^2 \cos^2 \theta} (2a)
y = x \tan \theta - \frac{gx^2 \sec^2 \theta}{2v^2} (2b) (Trigonometric identity)
y =x \tan \theta - \frac{gx^2}{2v^2}(1+ \tan^2 \theta) (2c) (Trigonometric identity)
0 = \frac{-gx^2}{2v^2} \tan^2 \theta + x \tan \theta - \frac{gx^2}{2v^2} - y (2d) (Algebra)
Let p = tanθ
0 = \frac{-gx^2}{2v^2} p^2 + xp - \frac{gx^2}{2v^2} - y (2e) (Substitution)
p = {\frac{-x\pm\sqrt{x^2-4(\frac{-gx^2}{2v^2})(\frac{-gx^2}{2v^2}-y)}}{2(\frac{-gx^2}{2v^2}) }} (2f) (Quadratic formula)
p = \frac{v^2\pm\sqrt{v^4-g(gx^2+2yv^2)}}{gx} (2f) (Algebra)
\tan \theta = \frac{v^2\pm\sqrt{v^4-g(gx^2+2yv^2)}}{gx} (2g) (Substitution)
\theta = \tan^{-1}{\left(\frac{v^2\pm\sqrt{v^4-g(gx^2+2yv^2)}}{gx}\right)} (2h) (Algebra)
Also, if instead of a coordinate (x,y) you're interested in hitting a target at distance r and angle of elevation φ (polar coordinates), use the relationships x = rcosφ and y = rsinφ and substitute to get:
\theta = \tan^{-1}{\left(\frac{v^2\pm\sqrt{v^4-g(gr^2\cos^2\phi+2v^2r\sin\phi )}}{gr\cos\phi}\right)}
No, I'm no engineer. I just mine their brains.
http://en.wikipedia.org/wiki/Trajectory_of_a_projectile
The total horizontal distance (d) traveled.
d = \frac{v \cos \theta}{g} \left( v \sin \theta + \sqrt{(v \sin \theta)^2 + 2gy_0} \right)
When the surface the object is launched from and is flying over is flat, the distance traveled is:
d = \frac{v^2 \sin(2 \theta)}{g}
As a special case, the distance is given by
d = \frac{v^2}{g}
when the angle (θ) is 45° and the initial height (y0) is 0.
For explicit derivations of these results, see Range of a projectile.
[edit] Time of flight
The time of flight (t) is the time it takes for the projectile to finish its trajectory.
t = \frac{d}{v \cos\theta} = \frac{v \sin \theta + \sqrt{(v \sin \theta)^2 + 2gy_0}}{g}
As above, this expression can be reduced to
t = \frac{\sqrt{2} \cdot v}{g}
if θ is 45° and y0 is 0.
The above results are found in Range of a projectile.
[edit] Angle of reach
The "angle of reach" (not quite a scientific term) is the angle (θ) at which a projectile must be launched in order to go a distance d, given the initial velocity v.
\sin(2\theta) = \frac{gd}{v^2}
\theta = \frac{1}{2} \arcsin \left( \frac{gd}{v^2} \right)
[edit] Conditions at an arbitrary distance x
[edit] Height at x
The height y of the projectile at distance x is given by
y = y_0 + x \tan \theta - \frac {gx^2}{2(v\cos\theta)^2} .
The third term is the deviation from traveling in a straight line.
[edit] Velocity at x
The magnitude, | v | , of the velocity of the projectile at distance x is given by
| v | = \sqrt{v^2 - 2gx \tan \theta + \left(\frac{gx}{v\cos \theta}\right)^2} .
[edit] Derivation
The magnitude |v| of the velocity is given by
| v | = \sqrt{V_x^2 + V_y^2} ,
where Vx and Vy are the instantaneous velocities in the x- and y-directions, respectively.
Here the x-velocity remains constant; it is always equal to v cos θ.
The y-velocity can be found using the formula
vf = vi + at
by setting vi = v sin θ, a = g, and t = \frac{x}{v \cos \theta}. (The latter is found by taking x = (v cos θ) t and solving for t.) Then,
V_y = v \sin \theta - \frac{gx}{v \cos \theta}
and
| v | = \sqrt{(v \cos \theta)^2 + \left(v \sin \theta - \frac{gx}{v \cos \theta} \right)^2} .
The formula above is found by simplifying.
[edit] Angle θ required to hit coordinate (x,y)
Vacuum trajectory of a projectile for different launch angles
To hit a target at range x and altitude y when fired from (0,0) and with initial velocity v the required angle(s) of launch θ are:
\theta = \tan^{-1}{\left(\frac{v^2\pm\sqrt{v^4-g(gx^2+2yv^2)}}{gx}\right)}
Each root of the equation corresponds to the two possible launch angles so long as both roots aren't imaginary, in which case the initial velocity is not great enough to reach the point (x,y) you have selected. The greatest feature of this formula is that it allows you to find the angle of launch needed without the restriction of y = 0.
Derivation
First, two elementary formulae are called upon relating to projectile motion:
x = v t \cos \theta , t = \frac{x}{v \cos \theta} (1)
y = vt \sin \theta - \frac{1}{2} g t^2 (2)
Solving (1) for t and substituting this expression in (2) gives:
y = x \tan \theta - \frac{gx^2}{2v^2 \cos^2 \theta} (2a)
y = x \tan \theta - \frac{gx^2 \sec^2 \theta}{2v^2} (2b) (Trigonometric identity)
y =x \tan \theta - \frac{gx^2}{2v^2}(1+ \tan^2 \theta) (2c) (Trigonometric identity)
0 = \frac{-gx^2}{2v^2} \tan^2 \theta + x \tan \theta - \frac{gx^2}{2v^2} - y (2d) (Algebra)
Let p = tanθ
0 = \frac{-gx^2}{2v^2} p^2 + xp - \frac{gx^2}{2v^2} - y (2e) (Substitution)
p = {\frac{-x\pm\sqrt{x^2-4(\frac{-gx^2}{2v^2})(\frac{-gx^2}{2v^2}-y)}}{2(\frac{-gx^2}{2v^2}) }} (2f) (Quadratic formula)
p = \frac{v^2\pm\sqrt{v^4-g(gx^2+2yv^2)}}{gx} (2f) (Algebra)
\tan \theta = \frac{v^2\pm\sqrt{v^4-g(gx^2+2yv^2)}}{gx} (2g) (Substitution)
\theta = \tan^{-1}{\left(\frac{v^2\pm\sqrt{v^4-g(gx^2+2yv^2)}}{gx}\right)} (2h) (Algebra)
Also, if instead of a coordinate (x,y) you're interested in hitting a target at distance r and angle of elevation φ (polar coordinates), use the relationships x = rcosφ and y = rsinφ and substitute to get:
\theta = \tan^{-1}{\left(\frac{v^2\pm\sqrt{v^4-g(gr^2\cos^2\phi+2v^2r\sin\phi )}}{gr\cos\phi}\right)}
No, I'm no engineer. I just mine their brains.
http://en.wikipedia.org/wiki/Trajectory_of_a_projectile
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LOBO Andy!!!!!!!!!!!!
That was great!
Glad everyone chimed in. With all of the crazy videos out there, it is obvious that many are fakes. That being said, they are still fun!
Sad part about it all, is that back in the day with the help of a little tequila (o.k. a lot of tequila) a bunch of us in college found the biggest hill we could find and went Ice Blocking. Now that was fun! Picture yourself atop a 50lb block of dry ice and a towel to keep your important parts from freezing and heading down a slope of about 35 degrees or so. Luckily no one was seriously injured (again thanks to the tequila). We tried that a few times along with broom ball (playing hockey with a huge red ball on an ice rink while wearing tennis shoes - yeah buddy!).
I wonder why my body protests so much these days......
:silly:
That was great!
Glad everyone chimed in. With all of the crazy videos out there, it is obvious that many are fakes. That being said, they are still fun!
Sad part about it all, is that back in the day with the help of a little tequila (o.k. a lot of tequila) a bunch of us in college found the biggest hill we could find and went Ice Blocking. Now that was fun! Picture yourself atop a 50lb block of dry ice and a towel to keep your important parts from freezing and heading down a slope of about 35 degrees or so. Luckily no one was seriously injured (again thanks to the tequila). We tried that a few times along with broom ball (playing hockey with a huge red ball on an ice rink while wearing tennis shoes - yeah buddy!).
I wonder why my body protests so much these days......
:silly: