during a softball game a batter hits a ball upward from an initial height of 3ft the height in feet of the softball is given by s(t)=-16t^2+80t+3 where t is time in seconds and t.0 which is closest to time when the softball will be 50 ft above the grpund
A 0.67 seconds or 5.04 seconds
B 2.50 seconds
C 0.83 seconds or 3.55 seconds
D5.04 seconds
To find the time when the softball will be 50 ft above the ground, we need to determine when the height of the softball, given by the equation s(t) = -16t^2 + 80t + 3, is equal to 50 ft.
Setting s(t) equal to 50 and solving for t:
50 = -16t^2 + 80t + 3
Rearranging the equation to form a quadratic equation:
-16t^2 + 80t - 47 = 0
Using the quadratic formula to solve for t:
t = (-b ± √(b^2 - 4ac)) / (2a)
where a = -16, b = 80, and c = -47:
t = (-80 ± √(80^2 - 4(-16)(-47))) / (2(-16))
t = (-80 ± √(6400 - 2992)) / (-32)
t = (-80 ± √3408) / (-32)
t = (-80 ± 58.37) / (-32)
Calculating both solutions:
t1 ≈ (-80 + 58.37) / (-32) ≈ 0.57 seconds
t2 ≈ (-80 - 58.37) / (-32) ≈ 5.42 seconds
Therefore, the closest time when the softball will be 50 ft above the ground is approximately 0.57 seconds. The correct option is not listed, so the closest option is C: 0.83 seconds or 3.55 seconds.
To determine the time when the softball will be 50 ft above the ground, we need to find the value of t for which s(t) = 50.
Given that s(t) = -16t^2 + 80t + 3, we can set it equal to 50 and solve for t:
-16t^2 + 80t + 3 = 50
Rearranging the equation, we have:
-16t^2 + 80t - 47 = 0
To solve this quadratic equation, we can either factor it or use the quadratic formula. In this case, the quadratic equation can be factored as:
(-8t + 1)(2t - 47) = 0
Setting each factor equal to zero, we get:
-8t + 1 = 0 or 2t - 47 = 0
Solving for t in each equation, we get:
-8t = -1 or 2t = 47
t = 1/8 or t = 47/2
Since time cannot be negative, we can disregard t = 1/8. Thus, the closest time when the softball will be 50 ft above the ground is t = 47/2, which is approximately t = 23.5 seconds.
Considering the options provided, the closest time is D) 5.04 seconds.
To find the time when the softball will be 50 feet above the ground, we need to solve the equation s(t) = 50.
The equation for the height of the softball is given as s(t) = -16t^2 + 80t + 3, where s(t) represents the height in feet and t represents time in seconds.
So, we substitute 50 for s(t) in the equation:
-16t^2 + 80t + 3 = 50
Rearranging the equation to get it in standard quadratic form:
-16t^2 + 80t + 3 - 50 = 0
-16t^2 + 80t - 47 = 0
Now, we need to solve this quadratic equation to find the values of t.
To solve the quadratic equation, we can either factor it or use the quadratic formula. In this case, the quadratic equation is not easily factorable, so we will use the quadratic formula.
The quadratic formula is given as:
t = (-b ± √(b^2 - 4ac)) / 2a
In our equation, a = -16, b = 80, and c = -47.
Now, let's substitute these values into the quadratic formula:
t = (-80 ± √(80^2 - 4(-16)(-47))) / (2(-16))
Simplifying further:
t = (-80 ± √(6400 - 2992)) / (-32)
t = (-80 ± √3408) / (-32)
Now, we calculate the square root of 3408:
t = (-80 ± 58.42) / (-32)
This gives us two possible solutions for t:
t = (-80 + 58.42) / (-32) ≈ 1.04 seconds
t = (-80 - 58.42) / (-32) ≈ 5.04 seconds
So, the two possible times when the softball will be 50 feet above the ground are approximately 1.04 seconds and 5.04 seconds.
Out of the options given, the closest time is 5.04 seconds. Therefore, the answer is D) 5.04 seconds.