Question

Could I have some quick help? will give brainliest if possible
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Answer 1

Answer:

a) 500 m

b) 10 s

c) 320 m

d) 8.94 s (2 dp)

Step-by-step explanation:

Given equation:   $\sf y=500-5x^2$

where:

• y = height of the raft above the water (in metres)
• x = time since the raft was dropped (in seconds)

a)  The height of the helicopter above the water is the height of the raft when x = 0:

$\sf \implies 500-5(0)^2=500 \ m$

b) The raft reaches the water when y = 0:

$\sf \implies 500-5x^2=0$

$\sf \implies 5x^2=500$

$\sf \implies x^2=100$

$\sf \implies x=\pm 10$

As time is positive, $\sf x=10 \ s$

c) The height of the raft above the water 6 s after it is dropped
is when x = 6:

$\sf \implies 500-5(6)^2=500 -180=320 \ m$

d) When the raft is 100 m above the water, y = 100:

$\sf \implies 500-5x^2=100$

$\sf \implies 5x^2=400$

$\sf \implies x^2=80$

$\sf \implies x=\pm \sqrt{80}$

As time is positive, $\sf x= \sqrt{80}=8.94 \ s \ (2 \ dp)$