From here you should be able to calculate the answer. One of them is -0.52 which is not the one you use. Minus heights do not matter in these problems yet. The other root is your answer.
Problem Two
Problem two is the usual way to express what happens to a projectile in the air. It again is going to be solved by the quadratic equation.
There are two ways to do this problem. If you know physics, you can solve it using the projectile formulas. If you don't , then you can use the trick in the previous problem. You can calculate how long it takes to hit the ground (h will then be zero).
This problem is rather nasty for someone who has not taken physics. There is another trick. You have to divide the time you get from the previous step by 2. When you solve the quadratic for h = 0, you get 11.6 seconds. But that is the time it takes to hit the ground. Half time is when the boulder will be at it's highest point. t = 11.6 / 2 = 5.8 (which your answers record as 5.75).
h = 0
a= -16
b = 184
c = 20
[tex]\text{x = }\dfrac{ -b \pm \sqrt{b^{2} - 4ac } }{2a} [/tex]
[tex]\text{x = }\dfrac{ -184 \pm \sqrt{184^{2} - 4*(-16)*(20) } }{2*(-16)} [/tex]
The answer from this equation is t = 11.6
The time you want is t = 5.75. You can calculate the maximum height by putting 5.75 in for t.
Problem 3
Problem 3 is done exactly the same way as problem two. You just have different numbers. When h = 0 the time taken to reach zero = 2.17 seconds. Therefore the time you want to put into the original equation for the maximum height is 1/2 of 2.17 seconds. From there you can find the maximum height.
Note: there is a lot of math in here. I have gotten you past the physics. I have also alluded to the answers. I think you should be able to carry out the rest.
