Posted by **Anonymous** on Monday, June 10, 2013 at 5:32pm.

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 ball’s height as a function of the horizontal distance it travels is given by . How far down the hill does the ball land?

- precalculus -
**Steve**, Monday, June 10, 2013 at 6:00pm
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 x-y 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

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