Question

the movement of the progress bar may be uneven because questions can be worth more or less (including zero) depending on your answer. 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 $h = -16t^2 + 70t + 4$ 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? \bigcirc 4.35 seconds \bigcirc 4.43 seconds \bigcirc 4.38 seconds \bigcirc 0.03 seconds

Explanation:

Step1: Set up the equation

We know the height equation is \( h = -16t^2 + 70t + 4 \), and we want to find \( t \) when \( h = 6 \). So we set up the equation:
\( -16t^2 + 70t + 4 = 6 \)
Subtract 6 from both sides to get a quadratic equation in standard form:
\( -16t^2 + 70t - 2 = 0 \)
We can multiply both sides by -1 to make the coefficient of \( t^2 \) positive:
\( 16t^2 - 70t + 2 = 0 \)

Step2: Use the quadratic formula

The quadratic formula is \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) for a quadratic equation \( ax^2 + bx + c = 0 \). Here, \( a = 16 \), \( b = -70 \), and \( c = 2 \).
First, calculate the discriminant \( D = b^2 - 4ac \):
\( D = (-70)^2 - 4 \times 16 \times 2 = 4900 - 128 = 4772 \)
Then, find the square root of the discriminant: \( \sqrt{4772} \approx 69.08 \)
Now, substitute into the quadratic formula:
\( t = \frac{70 \pm 69.08}{2 \times 16} \)
We have two solutions:
\( t_1 = \frac{70 + 69.08}{32} = \frac{139.08}{32} \approx 4.35 \)
\( t_2 = \frac{70 - 69.08}{32} = \frac{0.92}{32} \approx 0.03 \)
But we know the ball is caught after reaching maximum height, so we discard the smaller solution (0.03 seconds) because that's when it's going up. So we take \( t \approx 4.35 \) seconds? Wait, no, wait, let's check the calculation again. Wait, maybe I made a mistake in the sign. Wait, the quadratic equation after moving 6 to the left is \( -16t^2 + 70t - 2 = 0 \), so \( a = -16 \), \( b = 70 \), \( c = -2 \). Let's recalculate with \( a = -16 \), \( b = 70 \), \( c = -2 \).
Discriminant \( D = 70^2 - 4 \times (-16) \times (-2) = 4900 - 128 = 4772 \), same as before.
Then \( t = \frac{-70 \pm \sqrt{4772}}{2 \times (-16)} = \frac{-70 \pm 69.08}{-32} \)
Now, calculate the two solutions:
\( t_1 = \frac{-70 + 69.08}{-32} = \frac{-0.92}{-32} \approx 0.03 \)
\( t_2 = \frac{-70 - 69.08}{-32} = \frac{-139.08}{-32} \approx 4.35 \)
Wait, but the options are 4.35, 4.43, 4.38, 0.03. Wait, maybe my initial setup was wrong? Wait, the equation is \( h = -16t^2 + 70t + 4 \), set \( h = 6 \), so \( -16t^2 + 70t + 4 = 6 \), so \( -16t^2 + 70t - 2 = 0 \), multiply both sides by -1: \( 16t^2 - 70t + 2 = 0 \), then \( a = 16 \), \( b = -70 \), \( c = 2 \). Then \( t = \frac{70 \pm \sqrt{70^2 - 4*16*2}}{2*16} = \frac{70 \pm \sqrt{4900 - 128}}{32} = \frac{70 \pm \sqrt{4772}}{32} \). \( \sqrt{4772} \approx 69.08 \), so \( t = \frac{70 + 69.08}{32} \approx \frac{139.08}{32} \approx 4.35 \), or \( t = \frac{70 - 69.08}{32} \approx 0.03 \). But the problem says "after it gets to its maximum height", so we need the larger time. Wait, but let's check the maximum height time. The time to reach maximum height for a quadratic \( at^2 + bt + c \) is \( t = -b/(2a) \). For \( h = -16t^2 + 70t + 4 \), \( t = -70/(2*(-16)) = 70/32 \approx 2.19 \) seconds. So the ball is caught after 2.19 seconds, so we need the solution greater than 2.19. So between 4.35 and 0.03, 4.35 is greater than 2.19, so that's the solution. Wait, but the options have 4.35, 4.43, 4.38, 0.03. Wait, maybe I made a mistake in the quadratic formula. Let's use the original equation \( -16t^2 + 70t + 4 = 6 \), so \( -16t^2 + 70t - 2 = 0 \). Let's use the quadratic formula with \( a = -16 \), \( b = 70 \), \( c = -2 \). Then \( t = \frac{-70 \pm \sqrt{70^2 - 4*(-16)*(-2)}}{2*(-16)} = \frac{-70 \pm \sqrt{4900 - 128}}{-32} = \frac{-70 \pm 69.08}{-32} \). So \( t = \frac{-70 + 69.08}{-32} = \frac{-0.92}{-32} \approx 0.03 \), and \( t = \frac{-70 - 69.08}{-32} = \frac{-139.08}{-32} \approx 4.35 \). So the correct solution is approximately 4.35 seconds? Wait, but let's check with the equation. Let's plug t = 4.35 into \( h = -16t^2 + 70t + 4 \). \( t = 4.35 \), \( t^2 = 4.35^2 \approx 18.92 \), so \( -16*18.92 = -302.72 \), \( 70*4.35 = 304.5 \), so \( h = -302.72 + 304.5 + 4 = 5.78 \approx 6 \), which is close. So the answer is 4.35 seconds.

Answer:

4.35 seconds (Option: 4.35 seconds)