Answer :
The height is represented as y and the time is represented as x.
To find the maximum height (since this is a parabola that faces downwards) we need to find the vertex.
Vertex (x) = [tex] \frac{-(b)}{2a} [/tex]
b = 14
a = -16
Substitute
[tex] \frac{-14}{2(-16)} = \frac{-14}{-32} = \frac{7}{8} [/tex]
Now plug 7/8 to t
-16(7/8)^2 + 14(7/8)
-16(49/64) + 14(7/8)
-12.25 + 12.25
Answer: 0 ft
To find the maximum height (since this is a parabola that faces downwards) we need to find the vertex.
Vertex (x) = [tex] \frac{-(b)}{2a} [/tex]
b = 14
a = -16
Substitute
[tex] \frac{-14}{2(-16)} = \frac{-14}{-32} = \frac{7}{8} [/tex]
Now plug 7/8 to t
-16(7/8)^2 + 14(7/8)
-16(49/64) + 14(7/8)
-12.25 + 12.25
Answer: 0 ft
I suppouse the height is in meters and the time is in seconds.
1) we calculate the first derivative:
H´(t)=-32t+14
2) we math the first derivative to "0"; and we find out the value of "X".
-32t+14=0
t=-14/-32
t=7/16
3) we calculate the second derivative:
H´´(t)=-32<0; then at t=7/16 we have a maximum.
4) we find out the maximun height.
H(7/16)=-16(7/16)²+14(7/16)
H(7/16)=-3.0625+6.125
H(7/16)=3.0625≈3.063
Answer: the maximun height was 3.063 m.
If the height is in ft; then the answer is 3.063 ft.
1) we calculate the first derivative:
H´(t)=-32t+14
2) we math the first derivative to "0"; and we find out the value of "X".
-32t+14=0
t=-14/-32
t=7/16
3) we calculate the second derivative:
H´´(t)=-32<0; then at t=7/16 we have a maximum.
4) we find out the maximun height.
H(7/16)=-16(7/16)²+14(7/16)
H(7/16)=-3.0625+6.125
H(7/16)=3.0625≈3.063
Answer: the maximun height was 3.063 m.
If the height is in ft; then the answer is 3.063 ft.