An object is launched from ground level with an initial velocity of 120 meters per second. For how long is the object at or above 500 meters (rounded to the nearest second) ? The equation that models the path of the object is y = - 4.9^t2 + 120t.

A) 11 seconds

B) 12 seconds

Eliminate

C) 13 seconds

D) 14 seconds

Question

Jonah Mata

Mathematics

27 January, 10:13

An object is launched from ground level with an initial velocity of 120 meters per second. For how long is the object at or above 500 meters (rounded to the nearest second) ? The equation that models the path of the object is y = - 4.9^t2 + 120t.

A) 11 seconds

B) 12 seconds

Eliminate

C) 13 seconds

D) 14 seconds

+3

Answers (1)

Know the Answer?

Logan Hammond · Answer 1

To determine time in which the object is 500 meters above the ground, we just have to substitute 500 to y of the equation and determine the value of t.

y = - 4.9t² + 120t

Substituting,

500 = - 4.9t² + 120t

4.9t² - 120t + 500 = 0

The values of t in the equation is 5.32 and 19.17.