Re: Please check my calculations

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


From what you've posted so fan, I understand the bullet goes straight
up, and then what ... straight back down? If that's all there is to
the problem, there is no horizontal displacement. I have no good idea
where you got your equation for "displacement". Is there some other
aspect of this problem that hasn't been stated?

.

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