Projectile motion over an inclined plane

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MattDutra123
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Homework Statement
A projectile is launched perpendicular to a 30° slope at 20 m/s. Calculate the distance between the launching position and landing position.
Relevant Equations
Suvat equations of motion.
I have attached below the solution provided by the textbook. I understand the whole solution, except when they divide the horizontal component and the vertical component and equate it to the tangent of 30. What seems strange to me is that, while initially the equation for the vertical component was: "y = 20 × cos30° × t – 1/2 × 10 × t^2", when they equate it to the tangent they remove the "– 1/2 × 10 × t^2" section entirely. Is this a careless mistake by the book, or is there another reason for the removal of that section of the equation?
Thanks in advance.
 
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  • Diagram of a ball launched up from a 30 degree incline at 20 m/s, with an arc trajectory shown dashed, gravity of -10 m/s^2 marked downward, and horizontal distance x and vertical height y labelled, followed by worked equations for the components of motion.
    Screenshot 2019-06-24 at 16.01.48.png
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MattDutra123 said:
Problem Statement: A projectile is launched perpendicular to a 30° slope at 20 m/s. Calculate the distance between the launching position and landing position.
Relevant Equations: Suvat equations of motion.

I have attached below the solution provided by the textbook. I understand the whole solution, except when they divide the horizontal component and the vertical component and equate it to the tangent of 30. What seems strange to me is that, while initially the equation for the vertical component was: "y = 20 × cos30° × t – 1/2 × 10 × t^2", when they equate it to the tangent they remove the "– 1/2 × 10 × t^2" section entirely. Is this a careless mistake by the book, or is there another reason for the removal of that section of the equation?
Thanks in advance.
1/2 × 10 × t^2 is not removed.