Answer:
i.-34ft/s
ii.-27.6ft/s
iii.-26.8ft/s
iv-26.7ft/s
B.-26ft/s
Step-by-step explanation:
[tex]average velocity = \frac{y_{2}-y_{1}}{t_{2}-t_{1}}\\[/tex]
For all cases the value of the initial time [tex]t_{1} =2 seconds \\[/tex]
now we also use this time ti determine the value of the initial height in all cases
[tex]y_{1}=38(2)-16(2)^{2} \\y_{1}= 12ft\\[/tex]
now when the time is increase by 0.5 seconds the new height becomes
[tex]y_{2}=38(2.5)-16(2.5)^{2} \\y_{2}=-5fts\\[/tex]
the velocity becomes
[tex][tex]Average velocity = \frac{-5-12}{2.5-2}\\Average velocity = -34ft/s\\[/tex]
ii. when the time is increase by 0.1 seconds
the new height becomes
[tex]y_{2}=38(2.1)-16(2.1)^{2} \\y_{2}= 10.66ft\\[/tex]
[tex]Average velocity = \frac{10.66-12}{2.05-2}\\Average velocity = -27.6ft/s\\[/tex]
iii. when the time is increase by 0.05 seconds
the new height becomes
[tex]y_{2}=38(2.05)-16(2.05)^{2} \\y_{2}= 9.24ft\\[/tex]
[tex]Average velocity = \frac{9.24-12}{2.05-2}\\Average velocity = -26.8ft/s\\[/tex]
iv. when the time is increase by 0.1 seconds
the new height becomes
[tex]y_{2}=38(2.01)-16(2.01)^{2} \\y_{2}= 11.73ft\\[/tex]
[tex]Average velocity = \frac{11.73-12}{2.01-2}\\Average velocity = -26.7ft/s\\[/tex].
B. For us to determine the instantaneous velocity expression, we differentiate the expressing for the height
[tex]V_{inst}=38-32t\\[/tex]
we now substitute t=2, we arrive at
[tex]V_{inst}=-26ft/s\\[/tex]
