Question

A ball was thrown off a bridge. The table relates to the height of the ball above the ground in feet to the time in seconds after it was thrown. Use the data to enter a quadratic model in vertex form and convert it to standard form. Use the model to find the height of the ball at 1.5 seconds.​

Answer 1

The peak height reached at 144 feet, and the graph passing through points (0, 120), (1, 144), (2, 120), gives;

1. The vertex form of a quadratic equation is f(x) = a•(x - h)² + k

Where;

a = The leading coefficient

(h, k) = The coordinates of the vertex.

Let t represent the time of travel of the ball and let h represent the height.

From the given table, we have;

f(t) = a•(t - h)² + k

f(0) = 120 = a•(0 - h)² + k = a•h² + k

120 = a•h² + k

f(1) = 144 = a•(1 - h)² + k = a•(h² - 2•h + 1) + k

f(1) = a•(h² - 2•h + 1) + k

f(1) - f(0) = 144 - 120 = 24 = -((2•h - 1)•a

24 = -((2•h - 1)•a

f(2) = 120 = a•(2 - h)² + k

f(2) - f(1) = -24 = -((2•h - 3)•a

(f(1) - f(0)) - (f(2) - f(1)) = 48 = -2•a

Therefore;

a = -24

24 = -((2•h - 1)•a

24 = -((2•h - 1)×(-24)

-1 = -((2•h - 1)

2•h - 1 = 1

h = 2/2 = 1

h = 1

120 = a•(2 - h)² + k

120 = (-24)•(2 - 1)² + k

k = 120 + 24 = 144

• k = 144

Which gives;

The vertex form is y = -24•(x-1)² + 144

From the vertex form, we have;

y = -24•(x-1)² + 144 = -24•(x²-2•t+1) + 144

y = -24•(x²-2•tlx+1)+ 144 = -24•x²+48•x - 24 + 144

y = -24•x²+48•x + 120

y = -24•x² + 48•x - 24 + 144 = -24•x²+ 48•x + 120

y = -24•x² + 48•x + 120

The height of the ball after 1.5 seconds is

y = -24×1.5² + 48×1.5 + 120 = 138

• The height of the ball at t = 1.5 seconds is 138 feet

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