To solve the problem step-by-step:
Part A: Graph the function
To graph the function \( h(t) = -16t^2 + 40 \) using the Desmos calculator, go to Desmos and enter the equation. This will give you a parabola that opens downward since the coefficient of \( t^2 \) is negative.
Part B: Height of the school roof
To find the height of the school roof, we need to evaluate the function at \( t = 0 \) because this represents the initial height, or the height of the roof before the egg starts to fall.
\[ h(0) = -16(0)^2 + 40 = 40 \]
Thus, the point representing the height of the school roof is:
Ordered pair: \( (0, 40) \)
Part C: Height of the egg at \( t = 0.5 \) seconds
Now we calculate the height when \( t = 0.5 \):
\[ h(0.5) = -16(0.5)^2 + 40 \] \[ h(0.5) = -16(0.25) + 40 \] \[ h(0.5) = -4 + 40 = 36 \]
Therefore, the height of the egg at \( t = 0.5 \) seconds is 36 feet.
Part D: Height of the egg at \( t = 1 \) second
Now we calculate the height at \( t = 1 \):
\[ h(1) = -16(1)^2 + 40 \] \[ h(1) = -16 + 40 = 24 \]
Thus, the height of the egg at \( t = 1 \) second is 24 feet.
Part E: When does the egg hit the ground?
To find out when the egg hits the ground, we need to set \( h(t) = 0 \):
\[ -16t^2 + 40 = 0 \] \[ 16t^2 = 40 \] \[ t^2 = \frac{40}{16} = 2.5 \] \[ t = \sqrt{2.5} \approx 1.58 \]
Therefore, the egg hits the ground at approximately 1.58 seconds (rounded to two decimal places).
Summary of Answers
- Part B: (0, 40)
- Part C (0.5 seconds): 36 feet
- Part D (1 second): 24 feet
- Part E: 1.58 seconds