please please please someone help me no one on this site will help me. i so confused. please help.PLEASE SOMEONE HELP ME!!!!! ILL GIVE BRAINLIEST1. A model rocket is launched from a roof into a large field. The path of the rocket can be modeled by the equation y = –0.04x2 + 8.3x + 4.3, where x is the horizontal distance, in meters, from the starting point on the roof and y is the height, in meters, of the rocket above the ground. How far horizontally from its starting point will the rocket land?A. 208.02 mB. 416.30 mC. 0.52 mD. 208.19 m2. A catapult launches a boulder with an upward velocity of 184 ft/s. The height of the boulder (h) in feet after t seconds is given by the function h= -16t^2 + 184t + 20. How long does it take the boulder to reach its maximum height? What is the boulder's maximum height.A. Reaches a maximum height of 11.6 feet after 5.75 seconded.B. Reaches a maximum height of 549 feet after 11.5 seconds.C. Reaches a maximum height of 549 feet after 5.75 seconds.D. Reaches a maximum height of 23.2 feet after 11.6 seconds.3. A ball is thrown into the air with an upward velocity of 32 ft/s. Its height (h) in feet after t seconds is given by the function h= -16t^2 + 32t + 6. How long does it take the ball to reach its maximum height? What is the ball’s maximum height? Round to the nearest hundredth, if necessary.A. Reaches a maximum height of 22 feet after 1.00 second.B. Reaches a maximum height of 22 feet after 2.00 seconds.C. Reaches a maximum height of 44 feet after 2.17 seconds.D. Reaches a maximum height of 11 feet after 2.17 seconds.

Problem One
The rocket will hit ground level when y = 0. The plan is is to set y =0 and solve the quadratic. Be careful how you go about to this. It is an odd way to represent this kind of problem. The question does say that y defines how far off the ground the rocket is at the beginning. Therefore the height of the building does not matter (or shouldn't). All that matters is that y is ground level when y = 0.

a = - 0.04
b = 8.3
c = 4.3

[tex]\text{x = }\dfrac{ -b \pm \sqrt{b^{2} - 4ac } }{2a} [/tex]
[tex]\text{x = }\dfrac{ -8.3 \pm \sqrt{8.3^{2} - 4*(-0.04)*(4.3) } }{2(-0.04)}[/tex]
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 TwoProblem 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 = 0a= -16b = 184c = 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 3Problem 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.