Question

If a baseball player hits a baseball from 4 feet off the ground with an initial velocity of 64 feet per second, how long will it take the baseball to hit the ground? Use the equation h=-16t^2+64t+4.

Answer 1

Right now that equation is a function of time, h(t), height with respect to time.  If the baseball is on the ground, it has no height.  In other words, its height = 0.  So if we set the equation equal to 0 and solve for t, time, that will tell us the time that the ball had a height of 0.  If you plug those numbers into the quadratic formula, which is the best and most efficient way to factor a quadratic, you will get that the times are -.0615528128 and 4.061552813 seconds.  The 2 things in math that will never EVER be negative are time and distance/length.  So we know that the ball will not hit the ground at -.062 seconds.  Therefore, it hits the ground 4.06 seconds after it was hit.

Answer 2

Answer:

The answer is 4.06 seconds.

Step-by-step explanation:

1. We need to identify the value of h in the given equation. h stands for the final height of the ball. When the ball hits the ground, the height is 0, hence h=0, and now we replace

$0=-16t^{2} +64t+4$

2. We can see this  is quadratic equation, hence, we can use the quadratic equation to isolate the time " t". The quadratic formula is:

$\frac{-b+-\sqrt{b^2-4ac} }{2a}$

where "a" is the coefficient right next to t^2, "b" is the coefficient right next to t and "c" is the coefficient that is alone. For this problem, we have a=16, b=64 and c=4

3.  Now, we can replace

$t=\frac{-64+-\sqrt{64^2-4(-16)(4)} }{2(-16)} \\t=\frac{-64+-65.96 }{-32}$

Using the plus sign (+) we get t=-0.061 sec, and using the less sign (-) we get t=4.06 seconds.  We only need to take into account the positive time because there is no sense to have a negative time.