Taking the velocity of the distance formula, we get v = 16 - 1.6 T and taking the derivative of the velocity formula, we get acceleration equaling -1.6. So it says how long does it take the rock to reach its highest point? Well, its highest point is going to occur when the velocity is 0, because remember, if we're looking at this trajectory, the highest point comes in when the slope is 0. So we're going to take that velocity equation, set it equal to 0, and solve for time. So 0 equaling 16 -, 1.6 TT is going to equal 10. Then it says how high does the rock go? So we actually want to know the height, the S equation at time 10. So S equal, S of 10 equal. Substituting it in, we get 80. The next part says how long does it take the rock to reach half its height? So if we've thrown this rock and this is the height, we're going to have half the height here, but we're also going to have half the height here and have half the height on the way up and half the height on the way down. South half of 80 is 40 and we're going to set 40 equal to the height equation. Sixteen t -, .8 T squared. Taking everything to one side, I * 10 to get rid of the decimal. I realized it didn't factor, so then we put it into the quadratic formula and we get 17.07 and 2.93 approximately. Then it asked for how long is the rock aloft? So how long is it up in the air? Well, if we're here on the ground, we know the rocks on the ground at time zero and we want to know when else is the rock on the ground or the height is 0. So we take that that height formula again and now we're going to set it equal to 0. Realizing they each have AT, we could factor out the T. So t = 0 is when it's on the ground the first time, the 16 -, .8 T equaling 0 solving it's going to be on the ground the second time at T equal 20.