Re: Physics... math problem


"novelo" <novelo@xxxxxxxxx> wrote in message
news:1121219106.145516.311180@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
> Hello. I've got the following problem but I do not know if it
> solvable... I've been trying os far many methods, but still I cannot
> get it.
>
> Check is out:
>
> In a golf camp a hole is at 240ft horizonatlly and 64ft vertically. The
> ball hits the hole 4 seconds after it was hit.
> Get the magnitude and direction of the ball... Hence, get the Vo and
> the angle.
>
> + We cannot divide the time between 2 to know the maximun height
> because the altitude where you hit the ball is cero and the altitude of
> the hole is 64ft.
> + We cannot divide the distance in X either to get the highest point.
>
> Hence we cannot get the Vo of y neverthelles for X, Vo=64ft/s
>
> I've though to make this a parabola in the analytical geometry way...
> But we cannot get any info because the vertex, is unknown... (x,y) and
> therefore we cannot know the x-intercepts.
>
> I've tryed to do a relation, but I don't know if it will work...
>
> Since x=240 when y=64, when when y=0, x is?
> But the answer will be zero! And that cannot be that way! X keeps on
> growing forever while x is positive in a part and negative in the
> other...
>
> So... I think this problem is impossible to solve.
> If anyone has the answer or a key to get to it, that will be very
> helpfull.
> Maybe you think I am a little naive with this stupid high school
> problems, :P But it's kinda hard for me since I'm only about to enter
> high school and I've never seen this themes. I'm studying with books,
> so any help from you will really help me get more knowledge. Thank you
> once again!
>

Let's see if I remember how to do this :)

For the ball, we have x = v t (cos(theta)) and y = v t (sin(theta)) - 1/2 g
t ^2

We know the horizontal distance and the vertical distance so from the
equation for x, we have

240 = v ( 4 ) ( cos( theta ))

>From y, we get
y = v ( 4 ) (sin( theta )) - 1/2 g t ^2

Take y=64 ft, g=32 ft/s^2 and t = 4 sec
So we have y=4 v (sin(theta)) - 1/2 (32 ft/s^2) (16sec^2)

>From the equation for x, we can say v = 60/cos(theta). Substituting this
into the second equation gives

64 = 240( tan(theta))- 256
tan(theta) = 1.333
theta = 53 degrees
Therefore v=60/cos(53)=99.7 m/s

Dave


.

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