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+70t+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
To determine when the softball will be 50 feet above the ground, you can set the height function \( s(t) = -16t^2 + 70t + 3 \) equal to 50 and then solve for \( t \), the time in seconds.
So you want to solve the following quadratic equation:
\( -16t^2 + 70t + 3 = 50 \)
First, subtract 50 from both sides to set the equation to zero:
\( -16t^2 + 70t + 3 - 50 = 0 \)
\( -16t^2 + 70t - 47 = 0 \)
Next, you would typically use the quadratic formula to find the values of \( t \):
\( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
where \( a = -16 \), \( b = 70 \), and \( c = -47 \). Plugging in these values, you get:
\( t = \frac{-70 \pm \sqrt{70^2 - 4(-16)(-47)}}{2(-16)} \)
\( t = \frac{-70 \pm \sqrt{4900 - 4(16)(47)}}{-32} \)
\( t = \frac{-70 \pm \sqrt{4900 - 3008}}{-32} \)
\( t = \frac{-70 \pm \sqrt{1892}}{-32} \)
\( t = \frac{-70 \pm 43.48}{-32} \)
Now we have two possible solutions:
1. When adding the square root term:
\( t = \frac{-70 + 43.48}{-32} \)
\( t = \frac{-26.52}{-32} \)
\( t \approx 0.83 \) seconds
2. When subtracting the square root term:
\( t = \frac{-70 - 43.48}{-32} \)
\( t = \frac{-113.48}{-32} \)
\( t \approx 3.55 \) seconds
The softball will reach the height of 50 feet at approximately 0.83 seconds after being hit and again on its way down at approximately 3.55 seconds. Since the question asks for the time the ball will be 50 feet above the ground, the two closest times are approximately 0.83 seconds and 3.55 seconds after the ball is hit.