The equation y = -4.9t^2 + 23.8t describes the height (in meters) of a projectile launched from the ground at 23.8 meters per second. In how many seconds will the projectile first reach \(18\) meters in height?

Guest Feb 7, 2022

#1**+2 **

y = the height, so we can plug in 18 for y.

\(-4.9t^2 + 23.8t = 18\)

Then we turn this into a quadratic.

\(-4.9t^2 + 23.8t - 18 = 0\)

Then we can use the quadratic formula \(t = {-b \pm \sqrt{b^2-4ac} \over 2a}\), where a is the coefficient of t^2, b is the coefficient t, and c is the constant.

Plugging in the values we get: approximately 3.92 second

**the answer is 3.92004 :) **

proyaop Feb 7, 2022