## Verification of these sums?

For the discussion of math. Duh.

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HappySmileMan
Posts: 52
Joined: Fri Nov 09, 2007 11:46 pm UTC

### Verification of these sums?

Hey just wanted to ask if anyone can verify these sums I've done to do with projectiles.

A 181KG projectile is launched at 45 degrees after being accelerated with constant acceleration for 120KM, it travels 250KM before landing.
Calculate:
a) Initial Velocity
b) Time taken until landing.
c) Energy required.
d) Time required to accelerate projectile.
e) Acceleration.

a) Since it's being fired at 45 degrees I say it's launch at xi + xj (horizontal and vertical equal).

I let Sy = 0:
xt - 4.9t^2 = 0
t(x - 4.9t) = 0
t = 0s or t = x/4.9s

I now have the time taken in terms of x, from this I know that Sx = xt is:
x^2/4.9
Since it travels 250,000m I can say
x^2/4.9 = 250,000
x = sqrt(250,000 * 4.9)
x = 1,106.797m/s

Initial velocity would be x * sqrt(2) which is 1565.25m/s.

b) I know from previous part of question that t = x/4.9.
t = 1106.797 / 4.9 = 225.877s
(Or 250,000 / 1106.797 = 225.877)

c) Energy is 1/2*m*V^2
1/2 * 181 * (1565.25)^2 = 221725684.40625
= 221.726MJ

d) u = 0, v = 1565.25, s = 120,000
s = (u+v)/2 * t
120,000 = 782.625 * t
t = 153.330s

e) a = (v-u) / t
a = 1565.25/153.330
a = 10.21m/s^2

I've typed this out after writing it to a sheet so probably fairly messy and all over the place (and possibly has a few typoes), but can anyone confirm this is correct?

EDIT: Removed last 2 sentences since I got power and energy mixed up, and got confused between kW and W which really confused me about the result
Last edited by HappySmileMan on Tue Nov 11, 2008 9:39 pm UTC, edited 3 times in total.

Yakk
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### Re: Verification of these sums?

How to do your own double checking!

What is the kinetic energy of a 181KG object moving at 1565.25 m/s? You did htat.

What is the force needed to accelerate a 181KG object at 10.21m/s^2?

Over what distance is this force applied? Can you work out energy a different way?

When an object is at the top of it's arc, what is it's vertical velocity?

If an object has a K m/s vertical velocity, how high does it go? (use energy -- when does all kinetic energy be converted to gravitational potential energy?)

How long does it take that object to reach that peak?

It falls back down. When it is at the same height as when it started, how fast is it moving? (use energy)

How long will it take for it's speed to change from the speed at the peak, to the speed back at the same height? (use force)

An object moving at 1565.25 m/s at a 45 degree up angle -- what is it's vertical velocity? What is it's horizontal velocity?

Does it's horizontal velocity change?

How long before it hits the ground again, assuming a flat world?

How far does it go?

----

Basically, you derived the answers using some equations. Now take the results of what you derived, and solve for your original parameters, ideally using different routes through the physics.
Last edited by Yakk on Tue Nov 11, 2008 8:50 pm UTC, edited 1 time in total.
One of the painful things about our time is that those who feel certainty are stupid, and those with any imagination and understanding are filled with doubt and indecision - BR

Last edited by JHVH on Fri Oct 23, 4004 BCE 6:17 pm, edited 6 times in total.

HappySmileMan
Posts: 52
Joined: Fri Nov 09, 2007 11:46 pm UTC

### Re: Verification of these sums?

Yakk wrote:How to do your own double checking!

<snip>

Basically, you derived the equations. Now take the results of what you derived, and solve for your original parameters, ideally using different routes.

Hmm never really thought of checking it like that, doing it now and so far it all seems to be working out exactly like it did before.

Yakk
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Posts: 11129
Joined: Sat Jan 27, 2007 7:27 pm UTC
Location: E pur si muove

### Re: Verification of these sums?

This in particular:
1449kW energy over 2 and a half minutes

kW isn't energy, it's power. The power required isn't constant.

The force will be constant, but the distance it travels per unit time goes up.

Here is an additional problem to solve: what is the peak power output required for the launcher? (this might require calculus)
One of the painful things about our time is that those who feel certainty are stupid, and those with any imagination and understanding are filled with doubt and indecision - BR

Last edited by JHVH on Fri Oct 23, 4004 BCE 6:17 pm, edited 6 times in total.

HappySmileMan
Posts: 52
Joined: Fri Nov 09, 2007 11:46 pm UTC

### Re: Verification of these sums?

Yakk wrote:This in particular:
1449kW energy over 2 and a half minutes

kW isn't energy, it's power. The power required isn't constant.

The force will be constant, but the distance it travels per unit time goes up.

Here is an additional problem to solve: what is the peak power output required for the launcher? (this might require calculus)

Yes I typed before checking what I'd typed (Although I did just assume the power would be constant).

As for figuring out the peak power I just found that instantaneous power is Force by velocity, let F = 1848.01N (as I found earlier) and v = 1565.25m/s and got about 2.893MW.
No idea if that's right though or exactly how to check. Tried using Calculus to do it but couldn't really get anywhere (I haven't done all the calculus on my maths course anyway, so maybe I haven't come across something like this) so that's probably wrong, but it seems the best I could think of.

Also, nice avatar, I'm reading the 11th part of the comic series now.