Question

if a ball is thrown straight up into the air with an initial velocity of 8080 ft/s, its height in feet after tt second is given by y

Answer 1

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,  $y=80t-16t^2$ 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  $y=80t-16t^2$ .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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