Question

question 21 (mandatory) (1 point)
a disc is thrown into the air and follows a path modelled by the function ( h(t) = -5(t - 1.4)^2 + 10 ), where time, ( t ), is in seconds and height, ( h(t) ), is in metres. when does the disc hit the ground again?

a) 1.0 s

b) 2.8 s

c) 3.0 s

d) 1.4 s

Explanation:

Step1: Set height to zero

To find when the disc hits the ground, set \( h(t) = 0 \). So we have the equation:

$$ 0 = -5(t - 1.4)^2 + 10 $$

Step2: Solve for \( (t - 1.4)^2 \)

First, isolate \( (t - 1.4)^2 \). Add \( 5(t - 1.4)^2 \) to both sides:

$$ 5(t - 1.4)^2 = 10 $$

Then divide both sides by 5:

$$ (t - 1.4)^2 = \frac{10}{5} = 2 $$

Step3: Take square root

Take the square root of both sides. Remember, we consider the positive root for time (since time can't be negative in this context, and we want when it hits the ground again, so the larger time solution):

$$ t - 1.4 = \pm\sqrt{2} $$

We take the positive root (because the negative root would give a time before the disc was thrown, probably). So:

$$ t - 1.4 = \sqrt{2} \approx 1.414 $$

Step4: Solve for \( t \)

Add 1.4 to both sides:

$$ t \approx 1.4 + 1.414 = 2.814 \approx 2.8 $$

Answer:

b) 2.8 s