Re: Please check my calculations


Anon wrote:
On 12 Feb 2006 23:18:06 -0800, "Protoman" <Protoman2050@xxxxxxxxx>
wrote:


Anon wrote:
On 12 Feb 2006 21:06:03 -0800, "Protoman" <Protoman2050@xxxxxxxxx>
wrote:


Anon wrote:
On 12 Feb 2006 20:25:17 -0800, "Protoman" <Protoman2050@xxxxxxxxx>
wrote:

x is time in seconds, -16x^2 is the acceleration of gravity, 256x is
the initial velocity, f(x) is the equation for velocity, F(x) is the
equation for displacement, f'(x) is the equation for acceleration ,and
the units are feet and seconds. Hope this helps!!!

Not likely. Just post the statement of the problem that you are
trying to solve, OK?

I'm trying to solve this:

A bullet is launched from ground level at 180 degrees. The bullet has
an initial velocity of 256ft./sec. Find the max altitude, the time it
takes to get there, and the horizontal displacement at the max altitude.

Sorry I could not see your post, my news server apparently went down
for a couple of hours.

About your problem ... "launched from ground level at 180 degrees"?
What does that mean? Is that horizontal, or vertical, or something
else? Usually, 90 degrees means "straight up".

I meant 90 degrees. Sorry, I sort of think in upside down mirror
images, probably 'cause I'm left-handed.

OK, then basically your original results were correct. Taking
g = 32 ft/sec^2 and initial velocity v_0 = 256 ft/sec,

velocity = v(t) = -gt + v_0

At its highest point, the bullet's velocity is zero, which occurs
at time

32 t = 256
t = 8 seconds

bullet height = h(t) = -(1/2) g t^2 + v_0 t

Maximum height occurs at t = 8 seconds, so

maximum height = -(1/2) * 32 * 8^2 + 256 * 8 = 1024 feet

The results are right, but the equations are not.

f(x)=-16x^2+256x

If this is supposed to represent acceleration, it's not
correct. What is x? Time? The acceleration is constant.

This is the correct equation for height as a function
of time, if x is time.

f'(x)=-32x+256

This is the derivative of the above with respect to time
and is therefore the correct expression for velocity.

F(x)=-16x^3/3+128x^2+c

This is the integral of f(x) (which is usually the thing
given the name "vertical displacement" or just "displacement"
in your case, since the motion is entirely vertical).

- Randy

.

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