(ab/bc, non-calculator) the position function 2 st t ( ) 4.9 396.9   , gives the height (in meters) of an object that has fallen from a height of 396.9 meters after t seconds. (a) explain why there must exist a time t , 1 2  t  , at which the height of the object must be 382 meters above the ground. (b) find the time at which the object hits the ground. (c) find the average rate of change in s over the intervalt 8,9. include units of measure. explain why this is a good estimate of the velocity at which the object hits the ground. how can this estimate be improved? (d) evaluate: 3 ( ) (3) limt 3 st s  t   . show the work that leads to your answer. include units.

a)   y₁ = 392 m , y₂ = 377.3 m , t = 1.74 s ,   b)  v = 83.39.8 m/s , c) v = 88.2 m/s

Explanation:

Unfortunately some data of your statement is not clear, I will try to help you as much as possible.

The equation for free fall of a body is

        y = y₀ + v₀t -1/g  t²

As the body is released, its initial velocity is zero, the value of the acceleration of g = 9.8 m / s², the resulting equation is

       y = y₀ - 4.9 t²

a) From this equation we see that as time progresses the height decreases.

For t = 1s

        y = 396.9 -4.9 1²

        y₁ = 392 m

for t = 2 s

        y = 396.9 - 4.9 2²

        y₂ = 377.3 m

Let's find the time for the height of y = 382 m

         t = √ (y₀-y) 4.9

         t = √ [(396.9-382) / 4.9]

         t = 1.74 s

b) when the object hits the ground y = 0

       0 = y₀ - 4.9 t²

       t = √ y₀ / 4.9

       t = √ (396.9 / 4.9)

       t = 9 s

c) the rate of change of the position is the velocity,

t = 8 s

        y₈ = 396.9 - 4.9 8 2

        y₈ = 83.3 m

t = 9 s

        y₉= 396.9 -4.9 9 2

        y₉ = 0 m

the average exchange rate

        v = (x₈ -x₉) / Δt

       v = (83.8 -0) / 1

       v = 83.39.8 m / s

This is the average speed in the last interval so this average speed of this interval where it reaches the ground

c) The actual speed with which it reaches the ground

        v = dy / dt

        v = 2 (4.9) t

        v = 9.8 t

for t = 9s

        v = 9.8  9

        v = 88.2 m / s

How to improve these results using shorter time intervals

iron would move it is magnetic there is a mag field around the current in the wire

area of focus