marlena created a table of values representing the height in feet, ( h(t) ), of a coconut falling to the…

marlena created a table of values representing the height in feet, ( h(t) ), of a coconut falling to the hard ground with a height of 0 feet, collected over ( t ) seconds. assume the coconut does not bounce and rests on the ground when it lands. which statement is true? the initial height of the coconut is 190 feet. the coconut will hit the ground between 4 and 5 seconds after it was dropped. the values of ( h(t) ) when ( t = 4 ) and 5 should be 0. the maximum height of the coconut was 1 second after it was dropped. height of coconut over time

marlena created a table of values representing the height in feet, ( h(t) ), of a coconut falling to the hard ground with a height of 0 feet, collected over ( t ) seconds. assume the coconut does not bounce and rests on the ground when it lands. which statement is true? the initial height of the coconut is 190 feet. the coconut will hit the ground between 4 and 5 seconds after it was dropped. the values of ( h(t) ) when ( t = 4 ) and 5 should be 0. the maximum height of the coconut was 1 second after it was dropped. height of coconut over time

Answer

Explanation:

Step1: Analyze the initial height

When (t = 0), (h(0)=210) feet. So the initial height is 210 feet, not 190 feet.

Step2: Check when the coconut hits the ground

The coconut hits the ground when (h(t)=0). At (t = 3), (h(3) = 66); at (t=4), (h(4)=- 46). Since the function is continuous (assuming a smooth - falling motion), the zero - crossing (where (h(t) = 0)) occurs between (t = 3) and (t = 4) (using the Intermediate Value Theorem: if (y = h(t)) is continuous on ([a,b]) and (h(a)) and (h(b)) have opposite signs, then there exists a (c\in(a,b)) such that (h(c)=0)).

Step3: Check the values at (t = 4) and (t = 5)

The table shows (h(4)=-46) and (h(5)=-190). But since the coconut rests on the ground ((h(t) = 0) when it lands) and doesn't bounce, the negative values imply an error in data collection (maybe a miscalculation), but the coconut has already hit the ground before (t = 4).

Step4: Check the maximum height

The maximum height occurs at (t = 0) (initial height). At (t = 0), (h(0)=210); at (t = 1), (h(1)=194). Since (210>194), the maximum height is at (t = 0).

Answer:

None of the statements are completely correct. But if we assume some data - interpretation leniency (ignoring the negative - height data as a data - entry error):

  • The initial height statement is wrong ((h(0)=210\neq190)).
  • The “between (4) and (5)” statement is wrong (should be between (3) and (4)).
  • The “(t = 4) and (t = 5) should be (0)” is wrong (it should have hit the ground before (t = 4)).
  • The “maximum at (t = 1)” is wrong (maximum at (t = 0)).

If we consider the problem from a polynomial - fitting perspective (assuming (h(t)) is a quadratic function (h(t)=at^{2}+bt + c) with (c = 210) (from (t = 0)), (h(1)=a + b+210=194) (so (a + b=-16)), (h(2)=4a + 2b+210 = 146) (so (4a + 2b=-64)). Solving (\begin{cases}a + b=-16\4a + 2b=-64\end{cases}) (multiply the first equation by (2): (2a+2b=-32), subtract from the second: (2a=-32), (a=-16), then (b = 0)). So (h(t)=-16t^{2}+210). Set (h(t)=0): (-16t^{2}+210 = 0), (t^{2}=\frac{210}{16}=13.125), (t=\sqrt{13.125}\approx3.62) (still between (3) and (4)).

If we force - choose from the given options (despite inaccuracies in the problem's data presentation): The coconut will hit the ground between (4) and (5) seconds after it was dropped (if we wrongly consider the trend of the given (h(t)) values without proper function - fitting and error - analysis).