Posted by tracy on Thursday, October 20, 2011 at 11:17am.
OK. The four questions you posted are all basically the same. Here's help on this one:
The baseball will land when the height is zero.
So, you need to solve
-16t^2 + 50t + 4 = 0
Unless you see an abvikus factoring, use the quadratic formula:
for f(x) = ax^2 + bx + c,
x = (-b ± √(b^2 - 4ac)]/2
Here, a = -16
b = 50
c = 4
t = (-50 ± √(50^2 - 4(-16)(4))]/-32
= (-50 ± √(2500 + 256)]/-32
= (-50 ± √2756]/-32
= (-50 ± 52.5)]/-32
t = -0.07 or 3.2
Since time is positive, use t = 3.2 sec
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