Can someone help me with this I'm-- never mind I would just like help..
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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
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Let's assume that the coordinate system is defined as positive in the first quadrant.
The system has its origin in the place where Ingrid kicks the football ball.
We have then that the horizontal component is given by:
Rewriting we have:
Rounding the obtained result we have:
Answer:
The horizontal component of the initial velocity to the nearest tenth is: