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 will the ball be at 213ft from the ground?
At what will the ball be ... ????? Time ??
set -16t^(2)+128t = 213 and solve
16t^2 - 128t + 213 = 0
you will get two answers, one positive and one negative.
Reject the negative answer.
Ah, the ever-elusive physics question. Let's see if this clown can calculate the time it takes for that ball to be 213 feet from the ground.
We have the equation s(t) = -16t^2 + 128t, and we want to find when the ball is at 213 feet. So let's set s(t) equal to 213:
-16t^2 + 128t = 213
Now, we can rearrange the equation to make it a bit easier to solve:
16t^2 - 128t + 213 = 0
Using the magical powers of mathematics, we can solve this quadratic equation. Plugging it into an equation solver, we find that t ≈ 2.037 or t ≈ 6.963.
So, according to my calculations, the ball will be 213 feet from the ground at approximately t = 2.037 seconds and t = 6.963 seconds. Just make sure you have your umbrella ready when that ball comes falling down!
To find the time at which the ball will be 213ft from the ground, we need to solve the equation s(t) = 213.
Given:
s(t) = -16t^2 + 128t
Set s(t) = 213:
-16t^2 + 128t = 213
Rearrange the equation to solve for t:
-16t^2 + 128t - 213 = 0
Now we have a quadratic equation in the form ax^2 + bx + c = 0, where:
a = -16
b = 128
c = -213
We can solve this quadratic equation using the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / 2a
Substituting the values:
t = (-128 ± √(128^2 - 4*(-16)*(-213))) / (2*(-16))
Simplifying further:
t = (-128 ± √(16384 + 13632)) / (-32)
t = (-128 ± √(30016)) / (-32)
Now we can calculate the two possible values of t:
t1 = (-128 + √30016) / -32
t2 = (-128 - √30016) / -32
Calculating t1 and t2:
t1 ≈ 2.7 seconds
t2 ≈ 9.3 seconds
Therefore, the ball will be at 213ft from the ground at approximately 2.7 seconds and 9.3 seconds after it was projected upward.
To find the time at which the ball will be at 213 ft from the ground, we need to set up and solve the equation:
s(t) = -16t^2 + 128t
We want to find the value of t when s(t) is equal to 213 ft. So, we set up the equation:
-16t^2 + 128t = 213
To solve this quadratic equation, we need to rearrange it into the standard form, which is:
-16t^2 + 128t - 213 = 0
Now, there are different methods to solve this quadratic equation. One way is to use the quadratic formula, which states that for a quadratic equation ax^2 + bx + c = 0, the solutions can be found using the formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
For our equation, a = -16, b = 128, and c = -213. Plugging in these values into the quadratic formula:
t = (-128 ± √(128^2 - 4(-16)(-213))) / (2(-16))
Simplifying the equation further:
t = (-128 ± √(16384 + 13632)) / (-32)
t = (-128 ± √30016) / (-32)
t = (128 ± √30016) / 32
Now, we calculate the square root of 30016:
√30016 ≈ 173.219
Substituting this value into the equation:
t ≈ (128 ± 173.219) / 32
The two solutions are:
t ≈ (128 + 173.219) / 32 ≈ 301.219 / 32 ≈ 9.411
t ≈ (128 - 173.219) / 32 ≈ -45.219 / 32 ≈ -1.412 (discarding the negative solution as time cannot be negative)
Therefore, the ball will be at approximately 213 ft from the ground at around 9.411 seconds.