00:01
This problem, we're given this function for h of t, and it represents the height of an object after t seconds has been thrown.
00:07
So, part a, we want to find when the object can hit the ground.
00:10
Well, if the object hits the round, that means a tight would equal to zero.
00:14
So we're going to set h of t equal to zero.
00:16
So we'll have zero equal to one plus four t minus five t squared.
00:21
So now we just have to solve this equation.
00:23
So first off, i'm going to put it in standard form.
00:25
So we're going to have negative 5t squared plus 4t plus 1.
00:29
And then i'm going to divide both sides of our equation by negative 1 so that our first term will be positive.
00:36
So that means we'll have 0 equals 5t squared minus 4t minus 1.
00:42
Okay, so now we're going to solve this by factoring.
00:44
So i'm going to set up my two factors.
00:46
Now, the nice thing about our first term is that our leading coefficient is in prime number.
00:51
So we know our first terms will have to be 5t and t.
00:55
Now, our last term is equal to 1.
00:57
Well, the only things that multiply to 1 are 1.
00:59
1 and 1.
01:00
So now we just have to figure out which is positive and what's negative.
01:03
Well, because our middle term is negative, t minus 1 will have to be negative and the first one will be positive.
01:08
So that way we'll have negative 5t plus 1t.
01:11
Perfect.
01:12
So now that we have our factors, we'll set both of our factors equal to 0.
01:15
So 5t plus 1 could equal to 0 or t minus 1 could equal to 0.
01:22
And now we just solve both equations.
01:24
So to solve the first equation, we subtract 1 for both sides...