A rock is thrown from one side of a river to another. The function h (t)

inches of the rock t seconds after it has been thrown. (3 points)
a) Find h (2.5).
b) What does h (2.5) represent in the context of this problem?
c) Does it make sense to find h (-10) in the context of this problem?
:: 130
:: 30
:: 190
=
16t2 80t+ 30 gives the height in
2.5 seconds :: The time it takes for the height to be 2.5 inches above ground.
:: The height of the rock 2.5 seconds after it has been thrown.
:: The time it takes for the rock to hit the ground once it has been thrown.
:: Yes, it makes sense because the height could be under sea level.
:: No, it does not make sense because time cannot be negative.
:: No, it does not make sense because we cannot have negative inches.

1 answer

To answer your questions, let's break down each part:

Given Function:

The height of the rock as a function of time is given by: \[ h(t) = -16t^2 + 80t + 30 \] This is a quadratic function representing the height in inches of the rock at time \( t \) seconds after being thrown.

a) Find \( h(2.5) \):

Substituting \( t = 2.5 \) into the height function: \[ h(2.5) = -16(2.5)^2 + 80(2.5) + 30 \] Calculating each term:

  1. \( (2.5)^2 = 6.25 \)
  2. \( -16(6.25) = -100 \)
  3. \( 80(2.5) = 200 \)

Putting it all together: \[ h(2.5) = -100 + 200 + 30 = 130 \]

b) What does \( h(2.5) \) represent in the context of this problem?

In this context, \( h(2.5) = 130 \) represents the height of the rock, which is 130 inches above the ground, 2.5 seconds after it has been thrown.

c) Does it make sense to find \( h(-10) \) in the context of this problem?

Correct Answer: No, it does not make sense because time cannot be negative. Time \( t \) in this problem represents the duration after the rock has been thrown, and logically, you cannot have a negative time in this scenario.

Summary:

  • \( h(2.5) = 130 \)
  • It represents the height of the rock above the ground at 2.5 seconds.
  • Finding \( h(-10) \) does not make sense because negative time is not applicable in this context.