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A rocket is launched from a tower. The height of the rocket,

y in feet, is related to the time after launch, x in seconds, by the given equation. Using this equation, find out the time at which the rocket will land on the ground.

A rocket is launched from a tower. The height of the rocket, y in feet, is related-example-1

1 Answer

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The height of the rocket can be represented by the equation:

y = -3x^2 + 20x + 7

To find the time at which the rocket will land on the ground, we need to find the value of x when the height, y, equals zero. This is because the rocket will be on the ground when its height is zero.

So, we can write:

0 = -3x^2 + 20x + 7

To solve for x, we can use the quadratic formula:

x = [-b ± sqrt(b^2 - 4ac)] / 2a

In this case, a = -3, b = 20, and c = 7. Substituting these values into the quadratic formula, we get:

x = [-20 ± sqrt(20^2 - 4(-3)(7))] / 2(-3)

Simplifying this expression, we get:

x = [-20 ± sqrt(400 + 84)] / (-6)

x = [-20 ± sqrt(484)] / (-6)

x = [-20 ± 22] / (-6)

Therefore, the possible values of x are:

x = (-20 + 22) / (-6) = -1/3 or x = (-20 - 22) / (-6) = 7

The value of x = -1/3 is not a valid answer because time cannot be negative. Therefore, the rocketwill land on the ground after 7 seconds.

Thus, the time at which the rocket will land on the ground is:

7 seconds.
User YahyaE
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