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Westkost [7]
4 years ago
8

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 th

e given equation. Using this equation, find the time that the rocket will hit the ground, to the nearest 100th of second.
Mathematics
1 answer:
Mashutka [201]4 years ago
6 0

Here is the full question

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 the time that the rocket will hit the ground, to the nearest 100th of second.

Equation:

y=-16x^2+153x+98

Answer:

10.2 seconds

Step-by-step explanation:

Given equation :

y=-16x^2+153x+98 shown the expression of a quadratic equation

Let y =0

∴

0 = -16x^2+153x+98

where;

a=-16\\b=153\\c=98

Using the quadratic formula:

x=\frac{-b\±\sqrt{b^2-4ac}}{2a}

replacing it with our values; we have:

x=\frac{-153\±\sqrt{153^2-4(-16)(98)}}{2(-16)}

x=\frac{-153\ + \sqrt{153^2-4(-16)(98)}}{2(-16)}    OR     x=\frac{-153\ - \sqrt{153^2-4(-16)(98)}}{2(-16)}

x_1=10.17   OR     x_2=-0.60

Hence, we go by the positive value since,  is the time that the rocket will hit the ground.

x= 10.17

x = ≅ 10.2 seconds

Therefore, the rocket will hit the ground,  to the nearest 100th of second.  = 10.2 seconds

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