Question

You throw a ball upward. Its height h, in feet,
after t seconds can be modeled by the function
ℎ = − 16𝑡squared +30t+6. After how many seconds
will it hit the ground? (Use the Quadratic
Formula)

Answer 1

To find when the ball will hit the ground, we need to find the time (t) when the height (h) is equal to 0.

The given function is: h = -16t^2 + 30t + 6

Setting h equal to 0:
0 = -16t^2 + 30t + 6

Now we have a quadratic equation in the form ax^2 + bx + c = 0, with:
a = -16
b = 30
c = 6

Using the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)

Plugging in the values of a, b, and c into the quadratic formula:
t = (-(30) ± √((30)^2 - 4(-16)(6))) / (2(-16))

Simplifying:
t = (-30 ± √(900 + 384)) / (-32)
t = (-30 ± √1284) / (-32)

Calculating the square root and simplifying further:
t = (-30 ± 35.82) / (-32)

To find the two possible values of t, we have:
t1 = (-30 + 35.82) / (-32)
t2 = (-30 - 35.82) / (-32)

Evaluating:
t1 ≈ 0.66 seconds
t2 ≈ -1.41 seconds

Since time cannot be negative, we disregard t2 = -1.41 seconds.

Therefore, the ball will hit the ground approximately 0.66 seconds after it was thrown upward.