Whats the right answer
During a softball game, a batter hits a ball upward from an initial height of 3 feet. The height, in feet, of the softball is given by s left-parenthesis t right-parenthesis equals negative 16 t squared plus 70 t plus 3, where t is time in seconds and t greater than or equal to 0. Which is closest to the time when the softball will be 50 feet above the ground?
(1 point)
Responses
0.66 seconds
0.66 seconds
5.03 seconds
5.03 seconds
0.83 seconds or 3.55 seconds
0.83 seconds or 3.55 seconds
0.66 seconds or 5.03 seconds
To find the time when the softball will be 50 feet above the ground, we need to set the equation s(t) = 50 and solve for t.
-16t^2 + 70t + 3 = 50
-16t^2 + 70t - 47 = 0
We can solve this quadratic equation using the quadratic formula:
t = (-b ± √(b^2 - 4ac))/(2a)
Plugging in the values, we get:
t = (-70 ± √(70^2 - 4(-16)(-47)))/(2(-16))
Simplifying further:
t = (-70 ± √(4900 - 3016))/(-32)
t = (-70 ± √1884)/(-32)
Calculating the square root:
t = (-70 ± 43.42)/(-32)
Simplifying further:
t = (70 + 43.42)/32 OR t = (70 - 43.42)/32
t = 113.42/32 OR t = 26.58/32
t ≈ 3.55 seconds OR t ≈ 0.83 seconds
The closest answer to the time when the softball will be 50 feet above the ground is 0.83 seconds or 3.55 seconds.
To find the time when the softball is 50 feet above the ground, we need to solve the equation s(t) = 50.
Given the equation s(t) = -16t^2 + 70t + 3, we can set it equal to 50:
-16t^2 + 70t + 3 = 50
Bringing all the terms to one side, we get:
-16t^2 + 70t + 3 - 50 = 0
Simplifying,
-16t^2 + 70t - 47 = 0
To solve this quadratic equation, we can use the quadratic formula:
t = [-b +/- sqrt(b^2 - 4ac)] / (2a)
For this equation, a = -16, b = 70, and c = -47. Substituting these values into the quadratic formula, we can calculate the potential values for t.
t = [-70 +/- sqrt(70^2 - 4(-16)(-47))] / (2(-16))
t = [-70 +/- sqrt(4900 - 2992)] / (-32)
t = [-70 +/- sqrt(1908)] / (-32)
Since the quadratic equation has two potential solutions, we get two possibilities:
t1 = [-70 + sqrt(1908)] / (-32)
t2 = [-70 - sqrt(1908)] / (-32)
Calculating these values, we get:
t1 ≈ 0.83 seconds
t2 ≈ 3.55 seconds
Therefore, the closest time when the softball will be 50 feet above the ground is approximately 0.83 seconds or 3.55 seconds.
Thus, the correct answer is: 0.83 seconds or 3.55 seconds.
To find the time when the softball will be 50 feet above the ground, we need to solve the equation s(t) = 50.
Given that s(t) = -16t^2 + 70t + 3, we can set it equal to 50:
-16t^2 + 70t + 3 = 50
Next, we rearrange the equation to set it equal to zero:
-16t^2 + 70t + 3 - 50 = 0
This simplifies to:
-16t^2 + 70t - 47 = 0
To solve for t, we can use the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
In this equation, a = -16, b = 70, and c = -47.
Plugging these values into the formula, we get:
t = (-70 ± √(70^2 - 4(-16)(-47))) / (2(-16))
Simplifying further:
t = (-70 ± √(4900 - 2992)) / (-32)
t = (-70 ± √(1908)) / (-32)
At this point, we can calculate the values inside the square root:
t = (-70 ± √(1908)) / (-32)
t = (-70 ± √(2^2 x 3 x 7^2)) / (-32)
t = (-70 ± 14√3) / (-32)
Now, we have two possible values for t:
t1 = (-70 + 14√3) / (-32)
t2 = (-70 - 14√3) / (-32)
The only value close to the choices given is t1 ≈ 0.66 seconds.
Therefore, the closest time when the softball will be 50 feet above the ground is approximately 0.66 seconds. Hence, the correct answer is "0.66 seconds".