Question

The function H(t) = −16t2 + 64t + 12 shows the height H(t), in feet, of a baseball after t seconds. A second baseball moves in the air along a path represented by g(t) = 10 + 10.4t, where g(t) is the height, in feet, of the object from the ground at time t seconds.
Part A: Create a table using integers 1 through 4 for the 2 functions. Between what 2 seconds is the solution to H(t) = g(t) located? How do you know? (6 points)

Part B: Explain what the solution from Part A means in the context of the problem. (4 points)

Answer 1

Two functions are:
H ( t ) = - 16 t² + 64 t + 12
g ( t ) = 10 + 10.4 t
Part A:
--------------------------------------------
t ( sec )  |    1   |    2   |   3   |   4   |
--------------------------------------------
H ( t )     |  60   |  76   |  60  |  12  |
--------------------------------------------
g ( t )     | 20.4 | 30.8 | 41.2 | 51.6|
--------------------------------------------
The solution to H ( t ) = g ( t ) is located between 3 and 4 seconds.
We can check it :
- 16 t² + 64 t + 12 = 10 + 10.4 t
- 16 t² + 53.6 t + 2 = 0
When we solve it with the formula : t 1/2 = ( - b +/- sqrt( b² - 4 ac ) / 2a
solution is : t = 3.39 seconds.
Part B:
The solution from Part A shows when the second baseball becomes higher than the first baseball.

Answer 2

Answer:

Two functions are:

H(t) = -16t² + 64t + 12

g(t) = 10 + 10.4t

Part A:

to find: table showing value of h(t) and g(t) for t = 1 , 2 , 3 , 4

Table is attached.

The solution to H(t) = g(t) is located between 3 and 4 seconds. because between 3 & 4 value of H(t) goes from 60 to 12 and value g(t) goes from 41.2 to 51.6   only in these intervals we can get equal values of H(t) and g(t).

Part B:

Solution for t,

H(t) = g(t)

- 16t² + 64t + 12 = 10 + 10.4t

- 16t² + 53.6t + 2 = 0

We find value of t using quadratic formula,

$t=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

$t=\frac{-53.6\pm\sqrt{(53.6)^2-4\times(-16)\times2}}{2\times(-16)}$

we get,

t = 3.39 seconds other value is neglected because of negative value.

This shows that at time 3.39 seconds both balls are at same height.