precalculus
posted by Anonymous .
Berry is standing at the edge of a hill that slopes downward and to the right. He notices that for every 5 feet you go in a horizontal direction, the hill drops by 2 feet. Berry places a ball at the edge of the hill (the origin of the coordinate system), and kicks the ball at a speed of 20 m/s and at an angle of 30o to the horizontal. The equation describing the ballâ€™s height as a function of the horizontal distance it travels is given by . How far down the hill does the ball land?

the velocity has horizontal and vertical components
Vx = 20(√3/2) = 17.32
Vy = 20(1/2) = 10.00
The horizontal and vertical positions of the ball are thus
sx = 17.32t
sy = 10.00t  4.9t^2
So, the xy function of the height is
y = 10.00(x/17.32)  4.9(x/17.32)^2
= 0.01633x^2 + 0.5774x
The ground is descending with slope = 2/5 so the surface is given by
y = 2/5 x
So, where does the parabola intersect the line?
0.01633x^2 + 0.5774x = 0.4x
x = 59.85
y = 23.94
so the distance down the slope is
d = √(59.85^2 + 23.94^2) = 64.46 m