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 into 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 the top of the hill and y is the height, in meters, of the rock above the ground.

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 into 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 the top of the hill and y is the height, in meters, of the rock above the ground.
How far horizontally from its starting point will the rock land? round your answer to the nearest hundredth
56.00
24.54
57.04
57.26* my answer

Alex · Answer

can someone help?

Reiny · Answer

you must have a typo, since all your terms are positive.
Your rock will continue to rise higher and higher.
I was expecting the square term to be negative.

so I will assume that
-.04x^2 + 1.3x + 56 = 0 (y is zero when it hits the ground)
times -100
4x^2 - 130x - 5600 = 0
x = (130 ± √106500)/8
= 57.04 or some negative