Answer both questions and show your work

landon is standing in a hole that is 5.1 ft deep. he throws a rock, and it goes up into the air, out of the hole, and then lands on the ground above. the path of the rock can be modeled by the equation y = -0.005x^2 + 0.41x - 5.1, where x is the horizontal distance of the rock, in feet, from landon and y is the height, in feet, of the rock above the ground. how far horizontally from landon will the rock land?

a. 66.71 ft
b. 33.35 ft
c. 105.29 ft
d. 7.65 ft

a catapult launches a boulder with an upward velocity of 112 ft/s. the height of the boulder, h, in feet after t seconds is given by the function h = –16t^2 + 112t + 30. how long does it take the boulder to reach its maximum height? what is the boulder’s maximum height? round to the nearest hundredth, if necessary.

a. 7 s; 30 ft
b. 3.5 s; 366 ft
c. 3.5 s; 618 ft
d. 3.5 s; 226 ft

1. Since you're looking for the horizontal position of the rock when the rock is on the ground, you're looking for a solution where y=0 (since y is the vertical position of the rock). So we set y=0 and solve the equation, which gives us two results. One is about 15.3, the other about 66.7. Intuitively, we know that the rock will be at y=0 once on its way up, and then a second time on its way back down, when it lands. Since we're looking for its horizontal distance when it lands, we're looking for the larger solution, which is roughly 66.7 feet.

2.To determine the time it takes for the maximum height to be achieved, derive the  equation and equate to zero. 
                        h = -16t² + 112t + 30
                       dh / dt = 0 = 2(-16)t + 112
The value of t is 3.5 s. Substitute this value of time to the equation for h,
                         h = -16 x (3.5 s)² + 112 x 3.5 s + 30 = 226 ft
Thus, the answer is letter D.

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