Question

A water rocket is projected vertically upward from the ground. Its a height(h) m above the ground after t seconds is given by h=20-5(t-2)^2. When will the rocket reach the ground?
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Answer 1

Answer:

after 4 seconds

Step-by-step explanation:

Given

h = 20 - 5(t - 2)²

When rocket is on ground then h = 0

Equate to zero and solve for t

20 - 5(t - 2)² = 0 ( subtract 20 from both sides )

- 5(t - 2)² = - 20 ( divide both sides by - 5 )

(t - 2)² = 4 ( take the square root of both sides )

t - 2 = ± $\sqrt{4}$ = ± 2 ( add 2 to both sides )

t = 2 ± 2

Thus

t = 2 - 2 = 0 ← when rocket is launched

t = 2 + 2 = 4 ← when rocket hits the ground

The rocket reaches the ground after 4 seconds

Answer 2

The rocket will reach the ground 4seconds after.

Given the height reached by the rocket expressed as shown;

$h(t)=20-5(t-2)^2$

The rocket will reach the ground when h(t) = 0

Substitute h(t) = 0 into the expression as shown:

$h(t)=20-5(t-2)^2\\0=20-5(t-2)^2\\20-5(t-2)^2 = 0$

Expand the expression:

$20-5(t^2-4t+4) = 0\\20-5t^2+20t-20=0\\-5t^2+20t =0\\5t^2 = 20t\\5t = 20\\t = 20/5\\t=4secs$

This means that the rocket will reach the ground 4seconds after.

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