All of these equations will be set up as: h(t) = [tex]-\frac{g}{2}t^{2}[/tex] +vâ‚€t + hâ‚€ where g represents gravity, vâ‚€ represents initial velocity, and hâ‚€ 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 = [tex]\frac{-b}{2a}[/tex] = [tex]\frac{-160}{2(-16)}[/tex] = 5
Answer: 5 sec
1b) Â 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 Â Â Â Â Â Â Â 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 = [tex]\frac{-b}{2a}[/tex] = [tex]\frac{-288}{2(-16)}[/tex] = 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 Â Â Â Â Â Â Â t = 18
Answer: 18 sec
3e) Same concept as 1a
h(t) = -16t² + 352t
AOS: x = [tex]\frac{-b}{2a}[/tex] = [tex]\frac{-352}{2(-16)}[/tex] = 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 Â Â Â Â Â Â Â t = 22
Answer: 22 sec
