Rocket- Determining height and velocity from angle, distance travelled, and hangtime

[Thought Riot]

We shot a rocket.

Distance travelled: 982 feet
Angle shot at: 60*
Hang time: 20 seconds

Assuming that the rocket shot in a perfect parabola, how would you find out the height (apex) and velocity of the rocket?

Any help would be appreciated.

[Virus]

Well to find the apex, you'd just find the vertex of the parabola.

I don't like giving answers, but for the second you don't have enough info tbh. Air resistance, humidity, and all sorts of factors affect velocity. You could get a ball park figure using gravity. I'll let you take a stab at it and let everyone else chime in with their hints.

[Mantikore]

Actually, when it says it flies in a perfect parabola, it assumes theres no air resistance and all that

Displacement: 982 feet = 982*12*2.54/100 = 299.31 m lol metric
Angle: 60*
time: 20 s

so assuming
gravity = 9.8 m/s^2
no air resistance


anyway, you first need to find the initial velocity. the initial firing can be split into its horizontal and vertical components. what happens on the stuff acting horizontally doesnt effect whats happening vertically. Because there is no air resistance, there are no forces acting on the ball, so the horizontal velocity of the ball doesnt change at all during the flight.

so the time it took to travel 982 feet (299.31m) horizontally was 20 seconds
now since velocity = displacement/time
the horizontal velocity = 299.31/20 = 14.96 m/s

were in a good spot now since we can find the initial velocity. lets call the initial velocity v. remember you fired it at 60 degrees.

v * cos (60) = horizontal velocity................please ask if you dont understand this part. its important
=14.96.............rearrange to get v
v = 14.96/cos(60) = 29.93 m/s

now we find the initial vertical velocity. remember that unlike the horizontal component, there is a force affecting vertical flight (gravity).

v * sin (60) = initial vertical velocity
= 29.38* sin(60) = 25.92 m/s

now know that the velocity of an object at any point is affected by its initial velocity, its acceleration, and the time.

that is: v = u + a*t.............u is initial velocity, a is acceleration, t is time

since the displacement is the integral of velocity with respect to time

y = integral (u + a*t).....................y is the vertical displacement
=u*t + 0.5*a*t^2

if youre doing physics, you should have been told to remember these equations anyway. anyway, apply them since we have the initial vertical velocity, the acceleration (gravity) and we have the time. remember that the time at which it reaches its apex is half the time it took to travel that is (20/2 = 10 seconds)

so y = 25.92*10 + 0.5*9.8*10^2
=749.2 m
= 744*100/2.54/12 = 2458 feet
wait, is that right? some one check


edit: fuck, hold on, i think i made an error. you sure you got the details right, or not missing anything?

[BaconPie]

You can use calculus but you'll have to split it up into component vectors like the guy above said. Essentially you end up just using the equations.

I'd walk you through it but i'm not sure if you mean horizontal displacement or distance. They are two VERY different things.

The DISTANCE is the total length of the arc and the DISPLACEMENT is how far it went horizontally.

You're asking us for the Vertical Displacement.

Edit: Use SI units goddamn it!

[Mathematics]

This question is trivial, assuming a parabolic path is the same as neglecting air resistance. The horizontal speed and displacement then has nothing to do with the height reached and inital vertical speed.

The rocket reached its maximum height at 10 seconds. At this point it was stationary and then fell for 10 seconds under gravity before hitting the earth. The time is then given by:

h = a t^2 / 2 = 32.1 * 10^2 = 1605 feet (where 32.1 is the acceleration due to gravity in ft/s^2.

The inital speed is then given by solving:

1605 = u * t - a t^2 / 2

for u, known that the acceleration is 32.1 ft/s^2 and that t = 10.

This gives an inital speed in the vertical direction of 321 ft/s.

The inital velocity is then Sqrt(321^2 + (982/20)^2) = 324 ft/s at an angle of 60 degrees.

[harry_hardcore_hoedown]

Horitzontal velocity = horizontal distance / time = 299.3 / 20 ~ 15m/s (in SI).

Horizontal velocity is resultant velocity * the cosine of it's angle.

V = 15/cos 60 degrees

= 15/(1/2) = 30m/s

Vertical velocity = 30 sin 60 degrees
= 30(root3 / 2) ~ 26m/s

v = u + at
0 = 26 -9.8t
t = 26/9.8 = 2.65s

s = ut + 1/2at^2
s = 26(2.65) - 4.9(2.65)^2
s ~ 34.5m

Google tells me that that's a height of 113.19 feet, which doesn't seem very accurate, even with my questionable rounding. I must have fucked something up.

EDIT: I wonder why the time I calculated to reach maximum height is not even close to half it's flight time. It also seems to have started pretty slow by my calculations. Oh well, I see Mathematics worked it out.

[Mantikore]

The time is then given by:

h = a t^2 / 2 = 32.1 * 10^2 = 3210 feet (where 32.1 is the acceleration due to gravity in ft/s^2.


why didnt you divide it by 2? ie. shouldnt
h = a t^2 / 2 = 32.1 * 10^2 / 2 = 1605 feet??

[Mathematics]

why didnt you divide it by 2? ie. shouldnt
h = a t^2 / 2 = 32.1 * 10^2 / 2 = 1605 feet??

You're right. I'll edit my post.

Thanks.

[BoilingLeadBath]

"The problem is easy, assuming zero air resistance"

Yeah, but pointless.

Easy way to solve this is to brute-force search the two unknown variables in the launch: Cd and Velocity. You have enough known variables to do such a search, though results are liable to be fairly inaccurate. (Range is an awful correlate with Cd/V)

But I don't have a program written for this right now. It could be written in an hour or so... but I'm not that interested in your problem.

[Mathematics]

"The problem is easy, assuming zero air resistance"

Yeah, but pointless.

Easy way to solve this is to brute-force search the two unknown variables in the launch: Cd and Velocity. You have enough known variables to do such a search, though results are liable to be fairly inaccurate. (Range is an awful correlate with Cd/V)

But I don't have a program written for this right now. It could be written in an hour or so... but I'm not that interested in your problem.

What are you talking about?

[BaconPie]

What are you talking about?

He meant getting the answer by trial and error, only using a computer. It'd churn through all the possible values for the height until it found one that worked.

[harry_hardcore_hoedown]

He meant getting the answer by trial and error, only using a computer. It'd churn through all the possible values for the height until it found one that worked.

What's the point of that when you can work it out algebraically?

[Zip]

He meant getting the answer by trial and error, only using a computer. It'd churn through all the possible values for the height until it found one that worked.

Doesn't make sense in this problem anyway considering that it specifies a trajectory of a "perfect parabola" which wouldn't occur in more advanced models, and you would still need to know the mass density of the air and cross sectional area to model basic drag.

The point of these problems is to illustrate the concepts, not try to present a comprehensive or even realistic model. The solutions using the basic equations of motion are fine.

[The Jitterskull]

You're going to want an integral with einstein's equation to be truly accurate :mad:

[Mathematics]

You're going to want an integral with einstein's equation to be truly accurate :mad:

Modelling drag is a lot more important in obtaining an accurate answer, rather than an "integral with enstein's equation" (don't you mean a solution to the geodesic equation?). The difference between the Newtonian solution and a full GR treatment would be absolutely tiny.