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Vo=110 feet per second
h0=2 feet
So, h(t) = -16t^2 +110t +2
Take the derivative: h'(t) = 110 -32t
The maximum height will be at the inflection when the derivative crosses the x-axis aka when h'(t)=0.
So, set h'(t)=0 and solve for t:
0 = 110 -32t
-110 = -32t
t=3.4375
t=3.44 seconds
The maximum height, in feet, of the ball will attain 190.0 ft.
How to find the maximum height of the tennis ball machine?
Vertically into the air from a height of 2 feet
[tex]Y_{0}[/tex] = 2 feet
Initial height 110 feet per second
[tex]V_{0}[/tex] = 110 ft/s
[tex]Y_{max}[/tex] = ?
Formula:
[tex]V_{f} = V_{0} - g*t[/tex]
[tex]Y_{f} = Y_{0} + V_{0} *t - g*(t^2)/2[/tex]
1) [tex]Y_{max} = V_{f} = 0[/tex]
2) [tex]V_{f} =0[/tex] [tex]=V_{0} - g*t[/tex]
Where, t = [tex]V_{0}[/tex]/ g
[tex]g = 32.174 ft /s^2[/tex]
[tex]t = [110 ft/s ] / (32.174 ft/s^2)[/tex]
t = 3.419 s
3) [tex]Y_{f} = Y_{max}[/tex]
[tex]= 2ft + 110ft/s * 3.419s - (32.174 ft/s^2) (3.419s)^2 / 2[/tex]
= 190.0 ft
Therefore, the maximum height, in feet, of the ball will attain 190.0 ft.
To learn more about the maximum height
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