Question

A ball is dropped straight down from a height of 200 feet. After 1 second? The hall is 184 feet above the ground. After 2 seconds, the ball is 136 feet above the ground. Which equation models the height,y, of the hall x seconds after it’s dropped

Answer 1

The motion of the ball dropped from height is a free fall motion due to

gravitational acceleration.

Correct response;

• The equation that models the height is; y = -16·x²

Method for arriving at the above equation;

Given values;

$\begin{tabular}{|c|c|}Height (feet)&Time (s)\\200&0\\184&1\\136&2\end{array}\right]$

Required:

To select the equation that models the height, y, of the ball x seconds after

its dropped;

Solution:

From the above table, we have that the first difference is not a constant

The second difference is = 48 - 16 = 32

Taking the second difference as a constant, we have the following

quadratic sequence;

y = a·x²  + b·x + c

Where;

x = The time in seconds

y = The height after x seconds

• At x = 0, we have;

200 = a·0² + b × 0 + c

Therefore;

c = 0

• At x = 1, we have;

184 = a × 1² + b × 1 + 200

184 = a + b + 200

a + b = -16

a = -16 - b

• At x = 2, we have;

136 = a × 2² + b × 2 + 200

136 = 4·a + 2·b + 200

-64 = 4·a + 2×b

Therefore;

-64 = 4 × (-16 - b) + 2×b

-64 = -64 - 4·b + 2×b

b = 0

a + b = -16

Therefore;

a + 0 = -16

a = -16

The equation that models the height is; y = -16·x²

Learn more about quadratic function here;

brainly.com/question/2293136