Sunday, December 23, 2012

Finding the Velocity, Time and Distance for a Projectile Launched Above a Moving Target - Kind of Solved

a 1 cm projectile is launched at a 45° angle 100 m above a 1 cm target. the target is at a horizontal distance of 5 m from the vertical. just as the projectile is launched, the 1 cm target moves at a steady 9 m/s. there is no air resistance.

find the:

(1) velocity at which the projectile must be launched to hit the target
(2) maximum height of the projectile
(3) time it will take to hit the target
(4) horizontal distance where the hit will take place
(5) velocity of the projectile at the time of impact

Hx = initial horizontal distance of moving target
Hvx = velocity of moving target
v = velocity of projectile
v-impact = velocity at the time of impact
h = height projectile is launched from
α = angle
g = gravity at 9.8 m/s
t = time

we use the two equations which appear directly below the three sentences i'm about to type to find the velocity of the launched projectile. each equation represents the x-position of where the target and the projectile will meet. so far we have had only two sentences, but if i add another one we will have the three that i spoke of earlier in the first sentence of this paragraph. some would put a period after "two sentences" and then begin another sentence with but. however, that would make four sentences which is not the amount i promised there would be at the beginning. and anyway, it's too late now to go back and make the necessary changes (how many sentences have we had so far? i've lost count).

Hx + Hvx(t) = cosα v(t)

the following is the equation for time:

make an informed estimate, then do a trial and error.
sorry, but that’s the best i could do. hey, at least it works!

(if anyone can tell me how to algebraically incorporate the time equation into the above two equations to solve for velocity the way the good lord meant for it to be solved, i would appreciate it on a scale not seen in a fortnight)

plugging in the numbers we find:

v = 13.982 m/s
maximum height = 104.987 m
v-impact = 46.427 m/s 
t = 5.638 s
x = 55.739 m ≅ 55.7383 m

a difference of 0.7 mm. that’s pretty damn close for my liking.

maximum height = the top half of the first part of the equation for time.
v-impact = the square root of the square of Vx + Vy at the time of impact. Vx is cosα v and Vy can be found by multiplying the first part of the equation for time by 9.8.

Thursday, December 20, 2012

Kicking A Ball Into The Back Of A Moving Truck - Solved

A ball is kicked at an angle θ = 45°. It is intended that the ball lands in the back of a moving truck which has a trunk of length L = 2.5 m. If the initial horizontal distance from the back of the truck to the ball, at the instant of the kick, is 5 m, and the truck moves directly away from the ball at velocity V = 9 m/s, what is the maximum and minimum velocity (v) so that the ball lands in the trunk. Assume that the initial height of the ball is equal to the height of the ball at the instant it begins to enter the trunk.

Voy = original velocity of y
Vox = original velocity of x
t = time

this drove me crazy but i finally figured it out.

time that the ball will land is t = Voy/.5g or:

v sin45/4.9

(4.9 is 1/2g [g = gravity at 9.8 m/s])

the distance that the ball will be (distance-x) for minimum velocity is Voxt = 5 + 9t or:

v cos45(v sin45/4.9) = 5 + 9(v sin45/4.9)

distance-x for maximum velocity is Voxt = 2.5 + 5 +9t or:

v cos45(v sin45/4.9) = 2.5 + 5 + 9(v sin45/4.9)

we'll start with the distance equation to solve for the minimum velocity first. turn the motherfucker into a quadratic and solve for v this way:

5 + 9(v sin45/4.9) = v cos45(v sin45/4.9)

0 = 5 + 9(v sin45/4.9) - v cos45(v sin45/4.9)

let's put in the numbers we know:

0 = 5 + 9([v(.7071)]/4.9) - v(.7071) ([v(.7071)]/4.9)

v cos45 and v sin45 are the same.

now we can divide .7071 by 4.9 in both the first and second part of the equation:

0 = 5 + 9(v[.14431]) - v(.7071) (v[.14431])

in the first part we then have 9v multiplied by .14431, and in the second part we can multiply the 2 v's together as well as .7071 and .14431 to get:

0 = 5 + 9v(.14431) - v^2(.10204)

then we multiply 9v by .14431, and v^2 by .10204:

0 = 5 + 1.29879v - .10204v^2

here's our fucking quadratic.

let's rearrange it so it looks like this familiar cocksucker:

ax^2 + bx + c = 0

so that we have:

-.10204v^2 + 1.2988v + 5 = 0

then we can use the following formula:

to solve for v.

plugging in the numbers we find:

v = 15.82 m/s for minimum velocity

for the maximum velocity, just add 2.5 m to 5 m = 7.5 m

so then

v = 17.04 m/s


*note - for shits and giggles, try it with different starting angles and/or different starting positions and velocities for the truck. if you remember the equation for range time -  (v sin angle)/4.9 - you're in business.

Wednesday, December 19, 2012

Saturday, December 15, 2012

The Pee Velocity - Male Version

the other day i took a piss in the shower (actually, i left a piss in the shower rather than taking one) and i wondered about the velocity of the urine which left the "old chap".

if, for example, the pee began at a height of 1 meter and landed at a distance of 1 meter and took, let's say 0.2 seconds to hit the drain, then the velocity can be found in this way:

x-velocity = 5 m/s

if it went 1 meter in 0.2 seconds, then it will go 5 meters in 1 second. so the x-velocity is 5 m/s.

y-velocity = 4.02 m/s

if the urine began at a height of 1 meter and 0.2 seconds later it hit the drain, we can find the y-velocity with the following equation:

P (for position...not pee) = Voyt - 1/2gt^2 (t = time and g = gravity at 9.8 m/s...if i were pissing on the moon it would be a different story, all together...well come on! let's say it all together!)

plugging in the numbers we find:

-1 = Voy*0.2 - 0.196

after rearranging the equation to solve for Voy, we find the y-velocity is 4.02 m/s

v = 6.416 m/s

the velocity of the pee exiting das wienerschnitzel is the square root of x-velocity squared + y-velocity squared.

6.416 m/s. that's about 23 km/h. it would be like running the 100 m at the olympics in a time of 15.59 seconds. it's not a gold medal time....but it may be a golden shower medal time.

well, let's find the angle now shall we?

angle = 38.799 degrees below level

the angle of the dangle can be found by taking the -cos of dividing the x-velocity by the velocity, or by taking the -sin of dividing the y-velocity by the velocity.

or you can always just do the old arctan of rise over run. in other words the arctan of y-velocity/x-velocity.

so next time you're taking a whiz in the shower, i hope you appreciate the physics involved in such a simple every day act.

these calculations are also valid if you're pissing in the toilet.


the length of the piss stream can be approximated by dividing the time it takes to land into small segments, say for example 40, and then dividing the x-distance by the same amount. so we would have time segments of 0.004 seconds and x-distance segments of 0.02 m.

for each of the forty 0.02 x-segments we can calculate the distance that y would be after 0.004 seconds then add up all the little hypotenuses (or hypoteni if you prefer).

i took the liberty of baking one earlier and when i pull it out of the oven, we have an approximate stream length of 1.417 m.

now...on to a new pondering.

suppose we turned that 6.416 m/s velocity piss stream skyward. what's the maximum height we could achieve?

the maximum height would be if we aimed our urine straight toward heaven at a 90 degree angel...i mean angle. we could reach the dizzying height of 2.1 m. that's 2.1 m away from the willy and 3.1 m above the ground. holy doodlebugs!

at a 45 degree angle we could achieve maximum distance. here we would reach a height of 1.05 m, with a range of fully 5.034 m when it hits the ground. and a piss arc (not to be confused with the BC border crossing, the peace arch) of approx. 6.133 m that's a 4.82 m arc from a starting position of 1 m, up to a 1.05 m max. height, then back down to 1 m [where at this point it is 4.2 m away]. then from there, the stream will add another 1.313 m to its arc length until it finally and mercifully strikes the ground 5.034 m away).

just thought i'd share that with you.

tomorrow we'll discuss the tao of poo.

Tuesday, December 11, 2012

Home Run Conundrum - The Solution

A home run is hit in such a way that the baseball just clears a wall 21m high, located 130 m from home plate. The ball is hit at an angle of 35° to the horizontal, and air resistance is negligible.

(a) the initial speed of the ball,
(b) the time it takes the ball to reach the wall, and
(c) the velocity components and the speed of the ball when it reaches the wall. (Assume that the ball is hit at a height of 1.0m above the ground.)

Monday, December 10, 2012

Saguaro Construction

What Did the Busload of Non-Jews Sing When They Went on Vacation with Cliff Richard?

"We're all goyim on a summer holiday..."

Sunday, December 9, 2012

Stained Glass Saguaros

by dwdeclare

Psychedelic Saguaro

this picture was the first time i ever saw a saguaro. it's from april 2006 when i took a trip to the sonoran desert. and while i love all saguaros, this one will always be special. i've never seen another like it anywhere since.

Stained Glass Wire Saguaro

with inspiration from Léger, Malevich, and Kandinsky

Why Didn't the Jewish Candle Win the Vote to be Lit on Christmas Day?

because he was in the menorahty 
(just for the record, I sided with the katan yefeyfe ner - 
that means little beautiful candle, for all you goyim out there)

Saturday, December 8, 2012

Useful Equations for Projectile Motion

for any occasion
to use an equation
to figure how far a ball goes.
or maybe how high,
and the angle it flies
or velocity, now you will know.

so if we can take
an example and say,
when you multiply one half of g
with the square of the time,
the position of y
is the answer invariably

how easy it is
when the formulas give
the projectile motion you need.
just plug in the numbers,
there'll be no more blunders
in solving the problems you see.
RolandMoutal designed this using GeoGebra, I highly recommend his practical and informative creation for anyone who is interested in projectile motion. You can find it here.

Terms defined:

Vo = original velocity
α = angle
g = gravity (9.8m/s)
x = position of x along x-axis
y = position of y along y-axis
t = time
R = range max. or distance of x
h = height max. of y


to find the velocity of x when original velocity and angle are known:

Vox = Vo cos α

to find the velocity of y when original velocity and angle are known:

Voy = Vo sin α

to find the angle when the velocity of x and  y are known:


to find the angle when only the range and velocity are known: 


up to 45°, after that the equation becomes:


to find the angle when only height and velocity are known:

to find the position of x at a certain time when the velocity of x is known:

x = Voxt

to find the position of y at a certain time when the original velocity of y is known:

the velocity of x is found in the above equation (Vo cos α):

Vx =  Vox

to find the velocity of y at a certain time when the original velocity of y is known:

Vy = Voy – gt

max height for y can be found by:


to find velocity when the velocity of x and y are known:

(the pathagorean theorem a2 + b2 = c2)

to find the velocity when only the range and angle are known:

to find the velocity when only height and angle are known:

to find the time when y will be at a certain position (Δy) when that certain position and y’s original velocity are known:

Monday, December 3, 2012

Determining a Close Approximation of Parabolic Arc Length

...without all the mucky-muck of confusing calculus. maybe one day i will learn to comprehend and love the above formula for finding arc length, but for now i don't understand it (and i have tried to, oh lord, how i have tried) and i cannot bear to look at it. that integral symbol belongs on a violin...not in a math equation!

if only you could actually plug in some readily available numbers for dy and dx and have an answer pop out the other end....but noooooo, you have to have some prior esoteric knowledge of something else.

here's how i feel about it...


and now...on with an intuitively easier way

thank you to RolandMoutal on GeoGebra for creating a program which has allowed me to make this illustration. here is the actual interactive graphic

in the above example:

if an object has a velocity of 90 m/s at an angle of 37° then the velocity for x and y will be:

Vx = 90 cos 37 = 71.877 m/s
Vy = 90 sin 37 = 54.163 m/s

The maximum height of the object can be determined by using the following steps:


Velocity =  gt
where g = gravity 9.8m/s2 and t = time

plugging in the numbers we find:

54.163/9.8 = 5.527 seconds
the time it will achieve maximum height

then we use this formula to determine the position:

position = Voyt – ½gt2

54.163 (5.527) – 4.9 (30.546) = 149.68 m

what goes up, must come down at the same velocity and time it took to go up.
So it will take 5.527 seconds x 2 to go up and down = 11.054 seconds

to determine x-distance we multiply the x-velocity (71.877 m/s) times 11.054 seconds = 794.512 m

arc length:

for a close approximation of arc length, divide the x-distance (i.e., the range) and the time by a minimum of 20.
794.512 m/20 = 39.726 m
11.054 seconds/20 = .553 seconds

next determine the y-postion at those time intervals.
click for larger animated gif image
for example:

at .553 seconds (position 1 on the diagram below) the position of y will be 28.439 m
at 1.106 seconds (2 on the diagram) the position of y will be 53.884 m
at 1.659 seconds (3 on the diagram) the position of y will be 76.335 m
at 2.212 seconds (4 on the diagram) the position of y will be 95.793 m
and so on up to 5.527 seconds.

(you only need to do calculations for the first half of the arc, because the numbers for the second half are exactly the same...or as peter noone from herman's hermits might say, "second verse, same as the first! i'm a parabolic arc i am, parabolic arc i am, i am..." well, you know the rest)

The position of y will be determined by using the above formula for position.

Then we must find the length of each of those y-segments by subtracting the y-position below from the y-position above.
For example:

95.793 – 76.335 = 19.458 ( position 4 - 3 on the diagram)
76.335 – 53.884 = 22.451 (3 - 2)
53.884 – 28.439 = 25.445 (2 -1)
and the first length of y is 28.439 (position 1 on the diagram)

the x-axis segments will be constant at 39.726 m.

once you have determined all the y-lengths, you can then calculate the hypotenuse for each of the 10 little triangle segments of the first half of the arc by using the pythagorean theorem:

a2 + b2 = c2
or in our case:
x2 + y2 = hypotenuse

calculations for arc length when the range is divided by 20
for example:

√39.7262 + 28.4392 = 48.856m
√39.7262 + 25.4452 = 47.176m
√39.7262 + 22.4512 = 45.631m

and so on for all ten.

Once you have completed that, add up all ten and then multiply by 2 to achieve the complete length of the arc.

In our example, after all our calculations we should achieve an arc length of 864.182 m.
this is very close to the actual arc length of 864.327 we find by using the segment of a parabola calculator.

*note – to get even closer to the actual amount you can divide the x-distance and time into more segments.
Try 40, 50 or more (see below). But for all practical purposes, 20 works fine and will get you to within a few centimeters of the true arc length.
calculations for arc length when the range is divided by 40...getting warmer!

incidentally and fyi, the formula to find the area under the parabola is 2/3wh, i.e., width times height times .666. this is also affectionately known as the parabolic formula of the beast.

Friday, November 30, 2012

Swirl, Whirl, Twist and Twirl

Thursday, November 29, 2012




Wheel Construction in Blue and Red

Wednesday, November 28, 2012

Big Al's; Roaring 20s; Hungry I Club - SF

Tuesday, November 27, 2012

Monday, November 26, 2012

Musical Watch Dial