A ball is projected upward from the ground. Its distance in feet from the ground in t seconds is given by
s(t) = -16t^(2)+128t
At what times will the ball be at 213ft from the ground?
good grief! I just did this for you a few minutes ago!!!!
To find the times when the ball is at a specific height, we need to solve the equation for s(t) = 213. Let's substitute the given value into the equation:
-16t^2 + 128t = 213
To solve this quadratic equation, let's rearrange it into standard form:
-16t^2 + 128t - 213 = 0
Now, we can solve the equation by factoring, completing the square, or using the quadratic formula. In this case, let's use the quadratic formula:
The quadratic formula states:
t = (-b ± √(b^2 - 4ac)) / (2a)
For our equation, a = -16, b = 128, and c = -213. Substituting these values into the quadratic formula, we have:
t = (-128 ± √(128^2 - 4(-16)(-213))) / (2(-16))
Simplifying further:
t = (-128 ± √(16384 + 13632)) / -32
t = (-128 ± √(30016)) / -32
Now, let's calculate the square root of 30016:
√(30016) ≈ 173.06
Substituting this value into the equation:
t = (-128 ± 173.06) / -32
Now, we can calculate the two possible values of t:
t1 = (-128 + 173.06) / -32 ≈ 1.16
t2 = (-128 - 173.06) / -32 ≈ 10.53
So, the ball will be at 213ft from the ground approximately at t = 1.16 seconds and t = 10.53 seconds.