Question

A ball is thrown vertically upward from the ground with an initial velocity of 122 ft/sec. Use the quadratic function

Answer 1

Answer:

232.6 metres after 3.8 seconds.

Step-by-step explanation:

h(t) = -16t² + 122t

a = -16  b = 122  c = 0

Substitute into the quadratic formula

(Ignore the Â)

$x = \frac{-b±\sqrt{b^{2}-4ac}}{2a}$

$x = \frac{-122±\sqrt{122^{2}-4(-16)(0)}}{2(-16)}$

$x = \frac{-122±122}{-32}$

Split the equation at the ±

$x = \frac{-122+122}{-32}$        $x = \frac{-122-122}{-32}$

$x = \frac{0}{-32}$                    $x = \frac{-244}{-32}$

$x = 0$                                       $x = \frac{61}{8}$

The two x-intercepts at 0 and 61/8. The midpoint of the x-intercepts is the axis of symmetry, which is the x-coordinate of the vertex.

Midpoint = [0 + (61/8)] / 2

Midpoint = 61/16   <= This is the time for maximum height

t = 61/16

t = 3.8125  => Round to t = 3.8

To find the maximum height, substitute t=61/16 into the equation

h(t) = -16t² + 122t

h(61/16) = -16(61/16)² + 122(61/16)

h(61/16) = 232.5625  => Round to h = 232.6

Therefore, the ball will reach the maximum height of 232.6 metres after 3.8 seconds.