advanced algebra and trig.

B

b1955j1993

New member
Joined
Aug 19, 2009
Messages
10
help me help my daughter. a bullet is fired horizonal and the impact is heard in 1.5 sec. the bullet travel a 3300ft/sec. the speed of sound is 1100ft/sec. how do i go about getting an answer to help her
 
b1955j1993 said:
how do i go about getting an answer
Click to expand...


Hello b1995j1993:

You could start by telling us the question. 8-)

Is the exercise asking for the distance from the gun to the impact point?

It also helps tutors to have as much information about students' skill level as possible. In other words, it helps us to know what your daughter already knows about solving "distance equals rate times time" exercises like this one.

There are often different methods for approaching these exercises, as well. We won't have any way of knowing what methods your daughter is learning, unless somebody tells us.

Is your daughter familiar with solving a system of equations?

Cheers,

~ Mark

 
Hello, b1955j1993!

A bullet is fired horizonal and the impact is heard in 1.5 sec.
The bullet travel a 3300ft/sec. .The speed of sound is 1100ft/sec.
How do i go about getting an answer?
Click to expand...

But what is the question?
It would seem to be: "At what height was the bullet fired?"

The horizonral position of the bullet is given by: .\(\displaystyle x \;=\;3300t\)

The vertical position of the bullet is given by: .\(\displaystyle y \;=\;h - 16t^2\)
. . where \(\displaystyle h\) is the initial height of the bullet.


The bullet strikes the ground when \(\displaystyle y = 0\!:\)
. . \(\displaystyle h - 16t^2 \:=\:0 \quad\Rightarrow\quad t^2 \:=\:\frac{h}{16} \quad\Rightarrow\quad t \:=\:\frac{\sqrt{h}}{4}\)
\(\displaystyle \text{The bullet strikes the ground in: }\:t_1 \:=\:\frac{\sqrt{h}}{4}\text{ seconds.}\)


\(\displaystyle \text{In that time, its horizontal distance is: }\:x \:=\:3300\left(\tfrac{\sqrt{h}}{4}\right) \:=\:825\sqrt{h}\text{ feet.}\)

\(\displaystyle \text{To travel that distance, sound takes: }\:\frac{825\sqrt{h}}{1100}\text{ seconds.}\)
. . \(\displaystyle \text{Hence, the sound of impact reaches the shooter's ear in: }\:t_2 \:=\;\frac{3\sqrt{h}}{4}\text{ seconds.}\)


\(\displaystyle \text{The total time is 1.5 seconds: }\;t_1+t_2 \;=\;\frac{\sqrt{h}}{4} + \frac{3\sqrt{h}}{4} \:=\:1.5 \quad\Rightarrow\quad \sqrt{h} \:=\:1.5\)

\(\displaystyle \text{Therefore: }\;h \;=\;2.25\text{ feet.}\)

 


Here is one approach that requires us to disregard air resistance and the force of gravity. I have no idea whether or not these conditions are a part of this exercise, so I'm guessing here.

When the rate (speed) of a moving object (such as a bullet or a sound wave) remains constant (i.e., not speeding up or slowing down), then we have the following relationship between the distance traveled (d), the rate of travel (r), and the amount of elapsed travel time (t):

d = r * t

Here is the sequence of events, in this exercise (assuming that we are to disregard air resistance and the force of gravity).

At time t = 0, you simultaneously start a stopwatch and fire the gun.

Very soon, the bullet hits the target. At this impact point-in-time, the stopwatch shows some amount of elasped time (in seconds). We do not know this number of seconds. When values are unknown, we need to assign symbols to represent them.

X = the amount of elapsed time that it takes for the bullet to impact the target (in seconds)

As soon as the bullet hits the target, sound waves immediately start traveling from the target back to the gun. The stopwatch has been running continuously, so at the instant you hear the impact, the stopwatch again shows some amount of elasped time. We do not know how long it took the sound waves to reach your ear from the target, either. So, we assign a variable to represent this unknown number of seconds, too.

Y = the amount of elapsed time that it takes the sound waves to reach your ear (in seconds)

Of course, we have the following.

d = the distance from the gun to the target (in feet)

3300 = rate of the bullet (in feet/second)

1100 = rate of the sound waves (in feet/second)

We now have enough information to write equations d = r*t that model the senario.

The first equation below models the bullet, and the second equation models the sound waves

d = 3300X

d = 1100Y

The fact that the total amount of elapsed time (from the gun firing to hearing the impact) is 1.5 seconds is modeled by the following equation.

X + Y = 1.5

Since we have two different expressions for d, thes two expressions for d above must be equal to one another. Now we have a system of two equations.

3300X = 1100Y

X + Y = 1.5

Solve the system for X and Y. Then use either of these numbers in the corresponding equation for d above, to discover the distance in feet.

If there is anything that your daughter does not understand, then please try to get her to ask specific questions.

Cheers,

~ Mark
 
Thanks!! i'm sure with this info she should be able to answer the problem. you were very helpful and it looks like if she is to continue to make a's she will need lots of help from an outside source.
 
Use the formula distance = rate times time.

The bullet leaves the gun traveling at a rate of 3300 ft per second.
Let t represent the number of seconds the bullet is in flight.
The bullet hits the target making a sound.
The sound travels back toward the shooter at a rate of 1100 ft per second.
Total time from firing the fun to hearing the sound of impact is 1.5 seconds.
Let d represent the distance between gun and target.

(1) d = 3300t <<<Distance bullet travels = rate times time that the bullet is in flight.
(2) d = 1100(1.5-t) <<<Distance sound travels = rate times time the sound takes to get back to shooter.

Now, you have two equations in two unknowns. Hopefully, your daughter can solve for t and then for d.
 


Ah, I see that Soroban is only making one assumption, where I made two.

Soroban disregarded air resistance, but he did not disregard gravity.

Your daughter needs to think about these types of assumptions; they're usually stated in class or near the beginning of the relevant textbook chapter.

Hopefully, something in this thread will click.

 


Ah, I see that Loren and I are on the same track.

Loren made a substitution; I'll show you the corresponding substitution in the system of equations that I posted. (It's the first step in solving the system of two equations that I posted.)

3300X = 1100Y

X + Y = 1.5

Solve the second equation for Y.

Y = 1.5 - X

Substitute this expression for Y in the first equation.

3300X = 1100(1.5 - X)

Now, we have an equation containing only one variable. Solve it for X.

Substitute the result for X in the equation d = 3300X, to find d.

(Different ways to skin the same cat.) Good luck!

 
thanks!!!!!!!1111 this is a great site for help. i'm sure we use it a lot this year with this class advanced algebra and trig then next semeter dual pre cal.
 
You must log in or register to reply here.
Top