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:

Answer 1

A.) y = ax^2 + bx + c
46 = a(0)^2 + b(0) + c
c = 46

63 = a(1)^2 + b(1) + 46
a + b = 17 . . . (1)

48 = a(2)^2 + b(2) + 46
4a + 2b = 2
2a + b = 1 . . . (2)

1 = a(3)^2 + b(3) + 46
9a + 3b = -45
3a + b = -15 . . . (3)

(1) - (2) = (2) - (3) => -a = 16
a = -16
From (1), b = 17 + 16 = 33

The required model is y = -16t^2 + 33t + 46

b.) for t = 2.5: y = -16(2.5)^2 + 33(2.5) + 46 = -16(6.25) + 82.5 + 46 = -100 + 128.5 = 28.5
Therefore, the height of the ball at 2.5 seconds is 28.5 feet.

c.) At maximum height, dy/dt = 0
dy/dt = -32t + 33 = 0
t = 33/32 = 1.03125
Maximum height = -16(1.03125)^2 + 33(1.03125) + 46 = 63.015625 ≈ 63 feet.