Question

A projectile is launched into the air.

The function h(t) = -9.8t² + 14t + 78
gives the height, h, in meters, of
the projectile t seconds after it is
launched.
To the nearest thousandth of a
second, when will the projectile
land back on the ground?
(A) 1.429
B 2.196
C 2.821
3.625

Answer 1

To find when the projectile will land back on the ground, we need to determine the value of t when the height, h(t), is equal to zero.

The given equation for height is h(t) = -9.8t² + 14t + 78.

Setting h(t) to zero, we have:

0 = -9.8t² + 14t + 78.

Now, we can solve this quadratic equation to find the values of t.

Using the quadratic formula, which states that for an equation of the form ax² + bx + c = 0, the solutions are given by t = (-b ± √(b² - 4ac)) / (2a), we can substitute the values from our equation:

t = (-14 ± √(14² - 4(-9.8)(78))) / (2(-9.8)).

Simplifying further:

t = (-14 ± √(196 + 3057.6)) / (-19.6).

t = (-14 ± √(3253.6)) / (-19.6).

t = (-14 ± 57.05) / (-19.6).

Now we can calculate the two possible solutions:

1. t = (-14 + 57.05) / (-19.6) ≈ 2.821 (to the nearest thousandth of a second).

2. t = (-14 - 57.05) / (-19.6) ≈ -2.196 (ignoring the negative value since time cannot be negative in this context).

Therefore, the projectile will land back on the ground approximately 2.821 seconds after it is launched. So, the answer is option C.