Yo Nick! ( I know you love math problems)

[EdFred]

or anyone else who wants to give it a shot...
I was looking through files I had saved, and came across the Kitten Cannon, and I just got to thinking about projectiles and such, and came up with the following question:

If you were to launch a projectile at a 15° launch angle, at what velocity would you have to fire it, in order for it to go a distance of 100 meters across a flat area?
(Assume no wind resistance, no curvature of the earth, and the launch point is at ground level.)
Please show all work.

Bonus points for indicating how high it would be at its highest point.

What can I say, I was bored today.

 

[cherokeeflyboy]

How much does it weigh, shape?

 

[gismo]

How much does it weigh, shape?

Doesn't matter if you... "Assume no wind resistance, no curvature of the earth, and the launch point is at ground level."

 

[gismo]

The initial velocity (V1) has both horizontal and a vertical component. Vertical velocity (Vv1) = sin(15)*V1 and horiz. vel. (Vh1) = cos(15)*V1. Therefore Vv1=tan(15)*Vh1.

The horizontal velocity (Vh) is constant and the vertical (Vh) is decreasing at 1g (9.08m/sec^2). Vh will be 0 at the highest point of the trajectory which occurs at half the total horizontal distance or 50m.

The time to traverse halfway (THalf)= Vv1/g. THalf also equals 50m/Vh1. Substituting tan(15)*Vh1 for Vh1 gives 50m/Vh1 = tan(15)*Vh1/g. Solving for Vh1:

Vh1= (50*g/tan(15))^0.5 which is comes out to 41.1625m/sec.

V1 (muzzle velocity)= Vh1/cos(15) or 42.6146m/sec.

The max height (s) is g/2* Th^2 or Vv1^2/2g.

Since Vv1= sin(15) * V1, s = 6.6987m.

Edit: Oops, I thought the distance was 100 miles not 100 meters. Recalculated and re=posted.

 

[EdFred]

Close on the velocity
g = 9.8 not 9.08

here's how I went about it.

A = angle
V = Velocity
VsinA = Vertical component
VcosA= Horizontal component
t = time
g=-9.8

distance = velocity*time + acceleration*time2
d = Vt + at2
a = g

dv = VsinA*t + (g/2)t2

Our vertical vector (dv) from start to finish = 0
(datum plane to datum plane)

0 = VsinAt -4.9t2
factor out a t
0=VsinA-4.9t
4.9t = VsinA
t = VsinA/4.9

dh = VcosA*t
substitute for t
dh = VcosA*VsinA/4.9
dh = V2*sinA*cosA/4.9
490 = V2*sinA*cosA
490/*sinA*cosA = V2

V = sqrt(490/sinA*cosA)

V = 44.27188274


VsinA = 11.4584
VcosA = 42.763
t = 2.33845

for our vertical max

Vmax = VsinA*(t/2) + (g/2)(t/2)2

Vmax = 6.69873

 

[SkyHog]

I was just gonna answer the same thing that Lance did, but with the 9.8 Newtons in the place of the 9.08, but he beat me, so I feel I no longer need to.

 

[fgcason]

Close on the velocity
g = 9.8 not 9.08

9.8 Newtons

Units bad.
Always solve physics equations for units.
(Lance mangled the numbers but got units correct)

N = force
g = acceleration

N = (kg*m) / s²

a = m/s²

1g = 9.807 m/s² = 32.175 ft/s²

 

[SkyHog]

my bad - "Newtons per second" or "g"

Bah.

 

[gismo]

Close on the velocity
g = 9.8 not 9.08

Oops. Not sure how I made that mistake. I normally use 9.8 m/sec^2 as a "close enough" approximation for g, but I wanted to be more accurate for this and looked it up. Somehow I turned 9.807 into 9.08, don't ask me how. Using the correct answer for g I get 44.2877... which is pretty close to your answer using 9.8.

here's how I went about it.

A = angle
V = Velocity
VsinA = Vertical component
VcosA= Horizontal component
t = time
g=-9.8

distance = velocity*time + acceleration*time2
d = Vt + at2

You lost me there for a second by leaving out the "/2" (d= Vt + at2/2), but then you used the correct forumula below:

a = g

dv = VsinA*t + (g/2)t2
V = 44.27188274

Interestingly the value used for g has no effect on the height answer (I get 6.69783... with either value). If you look at my solution, you can see that g appears in the numerator and denominator of the final equation (look back at the source of V1 for the other 'g') so I guess that makes sense, but I wouldn't have guessed that. That means your projectile would reach the same altitude when launced at the velocity which gives the same lateral distance whether you performed the experiment on Earth or the Moon (ignoring air friction).

Vmax = 6.69873

 

[EdFred]

You know, I missed that too. Now that I look I see a 4.9 in the bottom and 490 in the top. So gravity has no effect on the shape of the parabola, but the time it takes is about 5.73 second (assuming the moons gravity is exactly 1/6th that of the Earth) vs the 2.33 seconds here.

The parabola is exactly the same. Hmmmm, I'm going to have to do some research on this.

And yeah, I missed a squared early in my explanation.

 

[deafsound]

wowsers

 

[SCCutler]

I knew all that.

 

[jesse]

I knew all that.

me too.

 

[kath]

My students will be getting a problem almost exactly like this in.... oh... about 1.5 weeks from today.

When it really gets fun is when you give them V and ask them to solve for the ANGLE. This throws 'em for a loop every time. It's a much harder problem because the final formula has both a sin(theta) and a cos(theta) in it, and they have to remember one of those obscure "double-angle" formulas in order to solve the problem.

(Classes start Tuesday, whoo hoo!)

--Kath

 

[EdFred]

My students will be getting a problem almost exactly like this in.... oh... about 1.5 weeks from today.

When it really gets fun is when you give them V and ask them to solve for the ANGLE. This throws 'em for a loop every time. It's a much harder problem because the final formula has both a sin(theta) and a cos(theta) in it, and they have to remember one of those obscure "double-angle" formulas in order to solve the problem.

(Classes start Tuesday, whoo hoo!)

--Kath

And keep in mind that there will be 2 correct answers. 45º +/- n

 

[Frank Browne]

me too.

Yep. Kids stuff. I just didn't want to show off.

 

[jkaduk]

I used to know how to do that, a long time ago. There is an equation from physics that I always try to keep in mind. KE=1/2MV^2. Which means that if I hit the ground in an airplane, It's gonna hurt.

 

[gismo]

I used to know how to do that, a long time ago. There is an equation from physics that I always try to keep in mind. KE=1/2MV^2. Which means that if I hit the ground in an airplane, It's gonna hurt.

Yeah, I can see you after a wing falls off the plane calculating the kinetic energy of the upcoming impact.