7. The Sandusky Little League uses a baseball throwing machine to help train 10-year old players to catch high pop-ups. It throws the baseball straight up with an initial velocity of 48 ft/sec from a height of 3.5 ft.
a. Find an equation that models the height of the ball t seconds after it is thrown.
b. What is the maximum height the baseball will reach? How many seconds will it take to reach that height?
8. Find the zeros of the function algebraically. f(x) = x2 + 2x – 8
9. Find the zeros of the function algebraically. f(x) = x2 + 2x + 8
10. Find the zeros of the function algebraically. f(x) = 2x2 – 10x + 3
-16x^2 + 48x + 3.5
How do you get the initial equation?
a. To find an equation that models the height of the ball t seconds after it is thrown, we can use the equation of motion for vertical motion:
h(t) = -16t^2 + v0t + h0
Where:
h(t) is the height at time t
v0 is the initial velocity (48 ft/sec)
h0 is the initial height (3.5 ft)
Plugging in the values, we get:
h(t) = -16t^2 + 48t + 3.5
b. To find the maximum height the baseball will reach, we can use the equation h(t) from part a. The maximum height occurs when the vertical velocity is zero. We can find this by taking the derivative of h(t) with respect to t and setting it equal to zero:
h'(t) = -32t + 48 = 0
Solving for t, we get:
-32t = -48
t = 1.5 seconds
So, it will take 1.5 seconds for the baseball to reach its maximum height. To find the maximum height, we can substitute this value back into the equation h(t):
h(1.5) = -16(1.5)^2 + 48(1.5) + 3.5
h(1.5) = 108.5 ft
Therefore, the maximum height the baseball will reach is 108.5 ft and it will take 1.5 seconds to reach that height.
For question 8, 9, and 10, to find the zeros of the function algebraically, we need to set the function equal to zero:
8. f(x) = x^2 + 2x - 8 = 0
To solve this quadratic equation, we can factor it:
(x - 2)(x + 4) = 0
Setting each factor equal to zero, we get:
x - 2 = 0 --> x = 2
x + 4 = 0 --> x = -4
So the zeros of the function f(x) are x = 2 and x = -4.
9. f(x) = x^2 + 2x + 8 = 0
Since this quadratic equation does not factor nicely, we can use the quadratic formula:
x = (-b ± √(b^2 - 4ac))/(2a)
Plugging in the values a = 1, b = 2, and c = 8, we get:
x = (-2 ± √(2^2 - 4(1)(8)))/(2(1))
x = (-2 ± √(-28))/2
Since we have a negative value inside the square root, this equation has no real solutions. Therefore, there are no zeros for the function f(x).
10. f(x) = 2x^2 - 10x + 3 = 0
Again, if this quadratic equation does not factor nicely, we can use the quadratic formula:
x = (-b ± √(b^2 - 4ac))/(2a)
Plugging in the values a = 2, b = -10, and c = 3, we get:
x = (-(-10) ± √((-10)^2 - 4(2)(3)))/(2(2))
x = (10 ± √(100 - 24))/4
x = (10 ± √76)/4
Simplifying further, we get:
x = (10 ± 2√19)/4
x = (5 ± √19)/2
So the zeros of the function f(x) are x = (5 + √19)/2 and x = (5 - √19)/2.