Question
Question
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(t)=−16t2+80t+3
, where t is time in seconds and t≥0
. Which is closest to the time when the softball will be 50 feet above the ground?(1 point)
Responses
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(t)=−16t2+80t+3
, where t is time in seconds and t≥0
. Which is closest to the time when the softball will be 50 feet above the ground?(1 point)
Responses
Answers
GPT 3.5
To find the time when the softball will be 50 feet above the ground, we need to set the height function equal to 50 and solve for t:
s(t) = -16t^2 + 80t + 3
Setting this equal to 50, we get:
-16t^2 + 80t + 3 = 50
Rearranging the equation:
-16t^2 + 80t - 47 = 0
To solve this quadratic equation, we can use the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / 2a
In this case, a = -16, b = 80, and c = -47. Plugging these values into the quadratic formula, we get:
t = (-80 ± √(80^2 - 4(-16)(-47))) / 2(-16)
Simplifying the equation:
t = (-80 ± √(6400 - 3016)) / (-32)
t = (-80 ± √(3384)) / (-32)
Since the problem specifies that t ≥ 0, we can ignore the negative value from the ± symbol. So, we have:
t = (-80 + √3384) / (-32)
Calculating the square root:
t = (-80 + √(4 * 846)) / (-32)
t = (-80 + 2√846) / (-32)
Approximating the value of square root of 846:
t ≈ (-80 + 2 * 29.09) / (-32)
t ≈ (-80 + 58.18) / (-32)
t ≈ -22.18 / (-32)
t ≈ 0.693 seconds
Therefore, the time when the softball will be 50 feet above the ground is closest to 0.693 seconds.
s(t) = -16t^2 + 80t + 3
Setting this equal to 50, we get:
-16t^2 + 80t + 3 = 50
Rearranging the equation:
-16t^2 + 80t - 47 = 0
To solve this quadratic equation, we can use the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / 2a
In this case, a = -16, b = 80, and c = -47. Plugging these values into the quadratic formula, we get:
t = (-80 ± √(80^2 - 4(-16)(-47))) / 2(-16)
Simplifying the equation:
t = (-80 ± √(6400 - 3016)) / (-32)
t = (-80 ± √(3384)) / (-32)
Since the problem specifies that t ≥ 0, we can ignore the negative value from the ± symbol. So, we have:
t = (-80 + √3384) / (-32)
Calculating the square root:
t = (-80 + √(4 * 846)) / (-32)
t = (-80 + 2√846) / (-32)
Approximating the value of square root of 846:
t ≈ (-80 + 2 * 29.09) / (-32)
t ≈ (-80 + 58.18) / (-32)
t ≈ -22.18 / (-32)
t ≈ 0.693 seconds
Therefore, the time when the softball will be 50 feet above the ground is closest to 0.693 seconds.
Related Questions
during a softball game a batter hits a ball upward from an initial height of 3ft the height in feet...
Whats the right answer
During a softball game, a batter hits a ball upward from an initial height o...
During a softball game, a batter hits a ball upward from an initial height of 3 feet. The height, in...