Answer:
<em>P=mgh</em>
<em>P=mghm=55</em>
<em>P=mghm=55g=9.8 or ~10</em>
<em>P=mghm=55g=9.8 or ~10h=27</em>
Explanation:
<em>hope</em><em> it</em><em> will</em><em> help</em><em> you</em><em> have</em><em> a</em><em> great</em><em> day</em><em> bye</em><em> and</em><em> Mark</em><em> brainlist</em><em> if</em><em> the</em><em> answer</em><em> is</em><em> correct</em>
<em></em>
<em> </em><em>#</em><em>c</em><em>arry </em><em>on </em><em>learning</em>
The average velocity for the time period beginning when t=1 and lasting
(i) 0.01 seconds = 63.84 ft/s
(ii) 0.001 seconds = 63.984 ft/s
Given that a ball is thrown with an initial velocity = 80 ft/s
Let 'y' be the height in feet after 't' seconds.
Given, gives the height in 't' seconds.
Average velocity = Rate of change of distance
= Change in distance/Change in time.
The initial time can be taken as 0 s.
When t =1 s, y = 80 - 16 = 64 ft
(1) t = 0.01 s
y = 80 x 0.01 - 16 x 0.01 x 0.01 = 0.7984 ft
Average velocity = (64 - 0.7984) / (1 -0.01) = 63.84 ft/s
(2) t = 0.001 s
y = 80 x 0.001 - 16 x 0.001 x 0.001 = 0.079984 ft
Average velocity = (64 - 0.079984) / (1 -0.001) = 63.984 ft/s
The question is incomplete. Find out the complete question below:
If a ball is thrown straight up into the air with an initial velocity of 80 ft/s, it height in feet after t second is given by .Find the average velocity for the time period beginning when t=1 and lasting
(i) 0.01 seconds
(ii) 0.001 seconds
Learn more about average velocity at brainly.com/question/6504879
#SPJ4