Question

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?
0.03 seconds
4.35 seconds
4.38 seconds
4.43 seconds

Explanation:

Step1: Substitute \(h = 6\) into the equation

We have the equation \(6=-16t^{2}+70t + 4\).
Rearrange it to the standard quadratic form \(ax^{2}+bx + c=0\). So, \(16t^{2}-70t + 2 = 0\). Here \(a = 16\), \(b=-70\), \(c = 2\).

Step2: Use the quadratic formula \(t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\)

First, calculate the discriminant \(\Delta=b^{2}-4ac=(-70)^{2}-4\times16\times2=4900 - 128=4772\).
Then \(t=\frac{70\pm\sqrt{4772}}{32}\).
\(\sqrt{4772}\approx69.08\).
We have two solutions for \(t\): \(t_{1}=\frac{70 + 69.08}{32}\approx4.35\) and \(t_{2}=\frac{70-69.08}{32}\approx0.03\).
Since we are looking for the time after the ball has reached its maximum height (we know that for a quadratic function \(y = ax^{2}+bx + c\) (\(a<0\)), the vertex occurs at \(t=-\frac{b}{2a}=-\frac{70}{2\times(-16)}=\frac{70}{32}\approx2.19\) seconds. So we take the larger value of \(t\)).

Answer:

4.35 seconds