Respuesta :
The equation for height as a function of time can be written as
.. h(t) = -16t^2 +110t +2
.. = -16(t^2 -55/8t) +2
.. = -16(t^2 -55/8t +(55/16)^2) +2 +55^2/16 . . . . . complete the square
.. = -16(t -55/16)^2 +191 1/16 . . . . . . . . . . . . . . . . . . vertex form
The maximum height the ball will reach is 191 1/16 feet.
.. h(t) = -16t^2 +110t +2
.. = -16(t^2 -55/8t) +2
.. = -16(t^2 -55/8t +(55/16)^2) +2 +55^2/16 . . . . . complete the square
.. = -16(t -55/16)^2 +191 1/16 . . . . . . . . . . . . . . . . . . vertex form
The maximum height the ball will reach is 191 1/16 feet.
Answer: 190.0 ft
Explanation:
Data:
- motion: vertical launch upward
- Yo = 2 feet
- Vo = 110 ft / s
- Y max = ?
Formula:
Vf = Vo - g*t
Yf = Yo + Vo*t - g*(t^2)/2
Solution
1) Ymax => Vf = 0
2) Vf = 0 = Vo - gt => t = Vo / g
g = 32.174 ft /s^2
t = [110 ft/s ] / (32.174 ft/s^2) = 3.419 s
3) Yf = Ymx = 2ft + 110ft/s * 3.419s - (32.174 ft/s^2) (3.419s)^2 / 2 = 190.0 ft
Answer: 190.0 ft
Explanation:
Data:
- motion: vertical launch upward
- Yo = 2 feet
- Vo = 110 ft / s
- Y max = ?
Formula:
Vf = Vo - g*t
Yf = Yo + Vo*t - g*(t^2)/2
Solution
1) Ymax => Vf = 0
2) Vf = 0 = Vo - gt => t = Vo / g
g = 32.174 ft /s^2
t = [110 ft/s ] / (32.174 ft/s^2) = 3.419 s
3) Yf = Ymx = 2ft + 110ft/s * 3.419s - (32.174 ft/s^2) (3.419s)^2 / 2 = 190.0 ft
Answer: 190.0 ft