The height of an object thrown upward from the floor of a canyon 106 ft deep, with an initial velocity of 120 ft per second, is given by the equation h=-16x^(2)+120x-106, where h represents the height of the object in feet after x second. How long will it take the object to rise to the height of the canyon wall?

Question

19-09-2019
Physics

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The height of an object thrown upward from the floor of a canyon 106 ft deep, with an initial velocity of 120 ft per second, is given by the equation h=-16x^(2)+120x-106, where h represents the height of the object in feet after x second. How long will it take the object to rise to the height of the canyon wall?

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Muscardinus Muscardinus · Answer 1 · 2019-09-23T11:23:21+02:00

Explanation:

The height of an object thrown upward from the floor of a canyon 106 ft deep, with an initial velocity of 120 ft per second. The equation is given by :

[tex]h=-16x^2+120x-106[/tex]

Since, the depth of the canyon is (-106 feet) and the time taken by the object to rise to the height of the canyon wall is calculated as :

h = 0

[tex]-16x^2+120x-106=0[/tex]

On solving the above quadratic equation,

x₁ = 1.023 seconds

and

x₂ = 6.477 seconds

So, the time taken by the object to rise to the height of the canyon wall is 1.023 seconds (ignoring 6.477 seconds). Hence, this is the required solution.