User:Izalithium/test

Section 6.3, Problem #42
Two-Softball Toss ''Chris and Linda warm up in the outfield by tossing softballs to each other. Suppose both tossed a ball at the same time from the same height, as illustrated in the figure. Find the minimum distance between the two balls and when this minimum distance occurs.'' To figure out this problem we must somehow plot the path of each softball, using their respective vectors (angles and speeds) to calculate where they will be the closest to one another. I have decided to denote Linda's feet as the origin. This way we don't deal with annoying negatives in our math: at Linda's hand it'd be harder to account for the ground, and starting anywhere on Chris would mean our whole graph would be in the negative quadrants. Linda's throw: x = f(t) (this is the horizontal distance) y = g(t) (this is the height of the ball) for interval t Є [, ] (this is as of yet undetermined) Linda's ball travels at a rate of 45 ft/sec at an angle of 44°. This can be resolved into a pair of x and y vectors. Let these vectors for Linda be known as x1 and y1. y1: sin(44°)=(y1/45) sin(44°)*45=y1 y1=31.25 x1: cos(44°)=(x1/45) cos(44°)*45=x1 x1=32.37 This means that x = 45cos(44°)*t and that y = -16t²+45sin(44°)*t+5 The x formula is essentially (45 ft/sec)*(45 degrees), since it is basically speed*angle. The y formula comes from the standard vertical formula (y= -16t²+V0*t+h0) where t = time, V0*t = initial velocity, and h0 = the initial height. We shall come back to Linda, as we need to compare her (especially her y formula) to Chris' throw.

Chris' Throw: x = f(t) (this is the horizontal distance) y = g(t) (this is the height of the ball) for interval t Є [, ] (this is as of yet undetermined) Chris' ball travels at a rate of 41 ft/sec at an angle of 39°. This can be resolved into a pair of x and y vectors. Let these vectors for Chris be known as x2 and y2. y2: sin(44°)=(y2/41) sin(39°)*41=y2 y2=25.8 x2: cos(44°)=(x2/41) cos(39°)*41=x2 x2=31.86 Taking into account that Chris is 78 feet away from Linda, this means that x = 78-41cos(39°)*t and that y = -16t²+41sin(39°)*t+5 The x formula is essentially (41 ft/sec)*(39 degrees), since it is basically speed*angle. The y formula comes from the standard vertical formula (y= -16t²+V0*t+h0) where t = time, V0*t = initial velocity, and h0 = the initial height.

Now that we have resolved these equations, we can use them to find the distance by way of the formula √[(x1-x2)²+(y1-y2)²] note: "[" and "]" refer to the extent of the square root symbol radical This becomes a monster equation of √[((45cos(44°)*t)-(78-41cos(39°)*t)²) + ((-16t²+45sin(44°)*t+5)-(-16t²+41sin(39°)*t)²)] which you can further resolve into √[((32.37*t)-(78-31.86*t)²) + ((-16t²+31.25*t+5)-(-16t²+25.8*t)²)] Extremely difficult work by hand, I used my graphing calculator to find the minimum value = (1.21, 6.6) see graph I This y value, 6.6, represents the height. This height is the lowest distance that will occur between the two softballs. Thus, the x value of 1.21 is essentially our t(ime) value and our answer, because it gives us the answer to our t intervals and now we can plug t in to our 2 sets of x and y equations, as I shall now do: note: more or less, the interval will = window for the interval t Є [0, 1.21]: Linda x = 45cos(44°)t = 39.17 Linda y = -16t²+45sin(44°)*t+5 = 19.47 Linda (x, y) = (39.17, 19.47)

Chris x = 78-41cos(39°)*t = 32.76 Chris y = -16t²+41sin(39°)*t+5 = 12.92 Chris (x, y) = (32.76, 12.92)

So, the answer to this problem really is that, at 1.21 seconds, the two softballs will be at the minimum distance from each other. But for the sake of thoroughness, I will plot their two paths as a graph (from which it can be guessed or mechanically calculated what their minimum value is) see graph II

Answer: The minimum distance between balls is achieved at 1.21 seconds.