Question

a physics student stands at the top of a hill that has an elevation of 56 meters. He throws a rock and it goes up in the air and then falls back past him and lands on the ground below the path of the rock can be modeled by the equation y=-0.04x^2+1.3x+56 where x is the horizontal distance in meters from the starting point on top of the hill and y is the height in meters of the rock above the ground. How far horizontally from the starting point will the rock land? Round to the nearest hundredth.

Answer 1

The answer is 57.04 meters. 
Solution:
The height y of the rock above the ground as a function of the horizontal distance x from the starting point on top of the hill can be modeled by the equation 
     y = -0.04x^2 + 1.3x + 56

We now set this equation equal to zero because the height of the ground where the rock will land is zero:
     0 = -0.04x^2 + 1.3x + 56

Since we have a quadratic equation, we can calculate for the horizontal distance x by using the quadratic formula     
     x = [-b ± sqrt(b^2 - 4ac)] / 2a      
     x = {-1.3 ± sqrt[1.3^2 - 4(-0.04)(56)]} / 2(-0.04)     
     x = (-1.3 ± sqrt(10.65)) / (-0.08)     
     x =  -24.54 meters or  57.04 meters
There are two solutions for x but one is negative, so we choose the positive value.