After how many seconds did the object reach its maximum height?

Question

The height in meters of an object thrown upward after t seconds is given by h(t )= -96t^2 +288t

Answer 1

h(t) = -96t² + 288t, where t is time (in seconds) and h is the height (in meters).

The maximum height refers to the vertex of the quadratic equation. We want to focus on time. We need to find the x-intercepts (solutions) and see what's in the middle of the two values.

Factor the quadratic using the greatest common factor (GCF).

-96t² + 288t = -(96t² - 288t) = -96(t² - 3t) = -96t(t - 3) (GCF = -96t)

Assume that the height is at zero and solve for t using zero product property.

-96t(t - 3) = 0 ⇒ -96t = 0 or t - 3 = 0 ⇒ t = 0 or t = 3

Add the two solutions and divide it by 2 to get t = (0 + 3) / 2 = 3/2 = 1.5 seconds.

The object will reach its maximum height at 1.5 seconds after it was thrown.