Question

PART B: APPLICATIONS

5. A Pokéball's flight is described by the equation h = –20t2 + 120t + 140, where h is the
height of the ball in feet at time, t seconds.
a) When does the ball hit the ground? (show all work)
[3]
b) From what height is the ball thrown?
[1]
c) What is the maximum height reached by the ball? (show all work)
[3]
PLEASE HELP AND SHOW WORK I BEG

Answer 1

Explanation:

the ball hits the ground when h=0

0=-20 t²+120 t+140

or t²-6t-7=0

t²-7t+t-7=0

t(t-7)+1(t-7)=0

(t-7)(t+1)=0

t=7,-1(rejected)

after 7 seconds the ball hits the ground.

when t=0,h=140

ball is thrown from a height of 140 ft

(c)

h=-20t²+120t+140

=-20(t²-6t+3²-3²)+140

=-20(t-3)²-20×-9+140

=-20(t-3)²+180+140

=-20(t-3)²+320

max. height is reached =320 ft

Answer 2

The ball hits the ground after 7 seconds.

The ball is thrown from a height of 140 ft.

The ball reaches a maximum height of 320 ft, which occurs at 3 seconds.

Explanation:

The easiest and fastest way to do this is to graph the equation in a graphing calc. When t = 0, we know that would be the y-intercept and we would find the initial height of the ball. The maximum vertex of the graph would be the maximum height of the ball (y-value) and where the graph crosses the x-axis would be where the ball hits the ground.