Question

The graph below shows the height of a projectile t seconds after it is launched. If acceleration due to gravity is –16 ft/s2, which equation models the height of the projectile correctly?

Answer 1

Answer:

  see attached

Step-by-step explanation:

The x-coordinate of the vertex is given by ...

  x = -b/(2a)

for f(x) = ax^2 +bx +c.

We see that the vertex is at x = 1, and we are given a = -16. Then we can find "b" from ...

  1 = -b/(2(-16))

  32 = b

We know the y-intercept of the equation is 5 (the starting height), so the equation must be ...

  h(t) = -16t^2 +32t +5 . . . . . matches the last choice

Answer 2

The quadratic equation which models the height of the projectile is - 16x² + 32x + c

The general form of a quadratic equation is :

• y = ax² + bx + c

From the graph ;

Intercept, c = 5

Acceleration due to gravity, a = - 16

x = vertex = 1

Using the relation, we can obtain the slope, b :

$x = \frac{ - b}{2a}$

$b = - 2ax$

b = - 2(-16)(1) = 32

Slope, b = 32

Therefore, the equation can be expressed thus :

• y = - 16x² + 32x + 5

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