a. The model of the quadratic equation is -16x² + 33x + 46
b. the height of the ball at 2.5 seconds is 28.5 fee
c. The ball's maximum height is 63.02
How to model the equation
a. The equation is modelled by using the standard quadratic equation formula and the given data
The quadratic equation formula
ax² + bx + c
From the table
at x = 0,
a(0)² + b(0) + c = 46, hence c = 46
at x = 1,
a(1)² + b(1) + 46 = 63
a + b = 17
at x = 2,
a(2)² + b(2) + 46 = 48
4a + 2b = 2
Solving the simultaneous equation for a and b
a + b = 17
4a + 2b = 2
a = 17 - b
4(17 - b) + 2b = 2
68 - 4b + 2b = 2
66 = 2b
b = 33
a = 17 - 33 = -16
hence a = -16 and b = 33
Substituting in to the quadratic equation is
-16x² + 33x + 46
b. At t = 2.5
= -16(2.5)² + 33(2.5) + 46
= 28.5 feet
c. maximum height is at x = -b/2a
= - 33/(2 * -16)
= 1.03
At t = 1.03
y = 16(1.03)² + 33(1.03) + 46
y = 63.02
complete question
A man throws a ball off the top of a building and records the height of the ball at different times as shown in the table:
Height of the ball
time (s) Height (feet)
0 46
1 63
2 48
3 1
a. Find a quadratic model for the data
b. Use the model to estimate the height of the ball at 2.5 seconds
c. What is the ball's maximum height?