Question

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

h(t) = -16t2 + 122t +0 to find how long it will take for the ball to reach its maximum height, and then find the
maximum height. Round your answers to the nearest tenth.

Answer 1

232.6 metres after 3.8 seconds.

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 =  (-122±122)/(-32)`

Split the equation at the ±

`x =  (-122+122)/(-32)`
`x =  (-122-122)/(-32)`

`x =  (0)/(-32)`
`x =  (-244)/(-32)`

`x = 0`
`x = (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.