Question

PLEASE HELP ME!!!! 60 POINTS!!!

Answer 1

Answer:

the first step is look at the answer above this one and copy that

Step-by-step explanation:

Answer 2

All of these equations will be set up as: h(t) = $-\frac{g}{2}t^{2}$ +v₀t + h₀ where g represents gravity, v₀ represents initial velocity, and h₀ represents initial height. When working with ft/sec, g = 32. So, -g/2 = -16

1a) Length of time to reach its maximum height means you are looking for the x-value of the vertex (aka Axis Of Symmetry).

h(t) = -16t² + 160t

AOS: x = $\frac{-b}{2a}$ = $\frac{-160}{2(-16)}$ = 5

Answer: 5 sec

1b)  Length of time to fall to the ground means you are looking for the x-intercept when height (y-value) = 0.

h(t) = -16t² + 160t

0 = -16t² + 160t

0 = -16t(t - 10)

0 = -16t        0 = t - 10

t = 0               t = 10

t = 0 is when it started, t = 10 is when fell to the ground.

Answer: 10 sec

2c) Same concept as 1a

h(t) = -16t² + 288t

AOS: x = $\frac{-b}{2a}$ = $\frac{-288}{2(-16)}$ = 9

Answer: 9 sec

2d) Same concept as 1b

h(t) = -16t² + 288t

0 = -16t² + 288t

0 = -16t(t - 18)

0 = -16t         0 = t - 18

t = 0              t = 18

Answer: 18 sec

3e) Same concept as 1a

h(t) = -16t² + 352t

AOS: x = $\frac{-b}{2a}$ = $\frac{-352}{2(-16)}$ = 11

Answer: 11 sec

3f) Same concept as 1b

h(t) = -16t² + 352t

0 = -16t² + 352t

0 = -16t(t - 22)

0 = -16t         0 = t - 22

t = 0              t = 22

Answer: 22 sec