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PostPosted: Sat, 16 Sep 2006 18:07:42 UTC 
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Joined: Sat, 16 Sep 2006 18:03:22 UTC
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I have these two problems for homework. I searched the forum for ways to do these problems and none yielded the correct answer. Please give me any help you can.

1. A gun is fired with angle of elevation 30°. What is the muzzle speed if the maximum height of the shell is 660 m?

2. A batter hits a baseball 3 ft above the ground toward the center filed fence, which is 9 ft high and 400 ft from home plate. The ball leaves the bat with speed 115 ft/s at an angle 50° above the horizontal. Is it a home run? (In other words, does the ball clear the fence?) Note: Use g = 32 ft/s2. How high is the ball above the ground when it reaches the fence?


Thanks!


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PostPosted: Sun, 17 Sep 2006 14:42:42 UTC 
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Joined: Mon, 19 May 2003 19:55:19 UTC
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Hello, Alethiometry!

You're expected to be familiar with "trajectory" problems.

With an angle of elevation of \theta, initial velocity v_o, and initial height h_o

. . the equations are: .x\:=\:(v_o\cos\theta)t\qquad y \:=\:h_o + (v_o\sin\theta)t - 4.9t^2


Quote:
1. A gun is fired with angle of elevation 30°.
What is the muzzle speed if the maximum height of the shell is 660 m?

This problem has: .\theta = 30^o,\;h_o = 0,\;v = muzzle speed.

The equations are: .x\:=\:\left(\frac{1}{2}v\right)t\qquad y \:= \:\left(\frac{\sqrt{3}}{2}v\right)t - 4.9t^2

Maximum height occurs when y' = 0.
. . y' \:=\:\frac{\sqrt{3}}{2}v - 9.8t \:= \:0\quad\Rightarrow\quad t = \frac{\sqrt{3}}{19.6}v seconds.

At the instant, the height is: .\left(\frac{\sqrt{3}}{2}v\right)\left(\frac{\sqrt{3}}{19.6}v\right) - 4.9\left(\frac{\sqrt{3}}{19.6}v\right)^2 \;= \;660

. . which simplifies to: .\frac{3}{78.4}v^2\:=\:660\quad\Rightarrow\quad v^2\:=\:\frac{78.4}{3}(660)\:=\:17,248


Therefore: .v\:=\:\sqrt{17,248} \:\approx\: \boxed{131.3\text{ m/sec}}



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 Post subject: Problem 2--homer or no?
PostPosted: Mon, 18 Sep 2006 16:39:07 UTC 
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yo = 3 ft
yf = ??
a= -g (-32.2 ft/s^2)
x = 400 ft
Vo = 115 ft/s
theta = 50 degree

Resolve Vo into its vector components:
Vox = Vo * cos 50 Voy = Vo * sin 50
Vox = (115 ft/s)(.6428) Voy = (115)(.766)
Vox = 73.92 ft/s Voy = 88.10 ft/s

x = Vox * t & yf = yo + Voyt - 1/2gt^2

t = x/Vox
t = 400 ft/73.92 ft/s
t = 5.41 s

Now, yf = 3ft + (88.10 ft/s)(5.41 s) - (16.1 ft/s^2)(5.41 s)^2
yf = 3 ft + 476.6 ft - 471.2 ft
yf = 8.40 ft

yf < 9 ft; therefore, ball does not clear fence. This is the general
gist for problems of this nature. You may want to check my
math.

MV

_________________
madvlad


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