+28 A baseball is hit with an initial upward velocity of 70 feet per second from a height of 4 feet above the ground. The equation h=−16t2+70t+4 models the height in feet t seconds after it is hit. After the ball gets to its maximum height, it comes down and is caught by another player at a height of 6 feet above the ground. About how long after it was hit does it get caught?
+9481 h = -16t2 + 70t + 4
We want to know how many seconds it takes for the ball to be 6 feet above the ground.
We want to know the value of t when h = 6 . So plug in 6 for h and solve for t .
6 = -16t2 + 70t + 4 Subtract 6 from both sides of the equation.
0 = -16t2 + 70t - 2 Let's solve this quadratic equation using the quadratic formula.
t = \(\frac{-70\pm\sqrt{70^2-4(-16)(-2)}}{2(-16)}\)
t = \(\frac{-70\pm\sqrt{4900-128}}{-32}\)
t = \(\frac{-70\pm\sqrt{4772}}{-32}\)
t = \(\frac{-70+\sqrt{4772}}{-32}\) ≈ 0.029 or t = \(\frac{-70-\sqrt{4772}}{-32}\) ≈ 4.346
So we know that
at 0.029 seconds, the ball is 6 feet above the ground and on its way up, and
at 4.346 seconds, the ball is 6 feet above the ground and on its way down.
The player will catch the ball on its way down, at about 4.346 seconds. 
+9481 h = -16t2 + 70t + 4
We want to know how many seconds it takes for the ball to be 6 feet above the ground.
We want to know the value of t when h = 6 . So plug in 6 for h and solve for t .
6 = -16t2 + 70t + 4 Subtract 6 from both sides of the equation.
0 = -16t2 + 70t - 2 Let's solve this quadratic equation using the quadratic formula.
t = \(\frac{-70\pm\sqrt{70^2-4(-16)(-2)}}{2(-16)}\)
t = \(\frac{-70\pm\sqrt{4900-128}}{-32}\)
t = \(\frac{-70\pm\sqrt{4772}}{-32}\)
t = \(\frac{-70+\sqrt{4772}}{-32}\) ≈ 0.029 or t = \(\frac{-70-\sqrt{4772}}{-32}\) ≈ 4.346
So we know that
at 0.029 seconds, the ball is 6 feet above the ground and on its way up, and
at 4.346 seconds, the ball is 6 feet above the ground and on its way down.
The player will catch the ball on its way down, at about 4.346 seconds.
