Helpful Hints for Unit 2 Test

Asymptotes and the TI Calculators

Asymptotes are the imaginary lines that represent a Domain value to which X can never be equal. Unfortunately, on the TI Calculator, these lines are drawn. Please ignore them.

Throwing things and the accompanying Mathematics

Whenever an object is thrown, it will follow a standard parabola, ignoring the effects of air resistance. When thrown all objects have three components:

1. Initial height: h_i

2. Initial speed (half of the velocity pair): s_i

3. Initial angle (the other half): a_i

With the effects of gravity, both height and distance can be calculated based on the time (t) of the flight. Given the initial velocity and height of the throw/launch of the projectile/baseball, the time until the ball reaches a particular height can be calculated. Once the time (t) is known, the distance can then be calculated.

Formulas

The Final Height (h_f) and distance (d_f) formulas are:

h_f = -16 ^feet/_second²*t² + s_i *sin(a_i)*t + h_i

d_f = s_i *cos(a_i)*t

where “*” stands for “multiplication” as is used in many programming languages.

Example:

If a baseball player throws a baseball at 150 ^feet/_second at an angle of 15^ofrom an initial height of 6 feet, how long would the ball be in the air before it hits ground and how far will it have traveled?

The formula I would first use would be:

h_f = -16 ^feet/_second²*t² + s_i *sin(a_i)*t + h_i

0 feet = -16 ^feet/_second²* t² + 150 ^feet/_second *sin(15^o)*t + 6 feet

Now, to make it easier, I would leave out the units and do all the mathematics needed to have a formula of at² + bt + c = 0 to be able to use the quadratic equation. Make sure the mode of your calculator is set to degrees and not radians before doing the mathematics.

-16t² + 150sin(15)t + 6 = 0

-16t² + 38.8t + 6 = 0

Now use the quadratic equation to solve for time (t).

……….

t = -b ± \/b² – 4ac

2a 2a

It would be best to solve parts of the equation before solving the entire equation:

………………..

t = -38.8 ± \/1505 – 4(-16)(6)

-32 -32

t = 1.2 ± 1.4 = 2.6 seconds or –0.2 seconds

Since we know that time does not travel backwards, only the 2.6 answer would be correct. Now use the final distance formula to determine the distance thrown:

d_f = s_i *cos(a_i)*t

d_f = 150 ^feet/_second *cos(15^o)*2.6 seconds

d_f = 150*.966*2.6 = 376.7 feet