Calculating Distance for Jumping off a Slope at 45°

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To calculate the distance a jumper travels when launching off a slope at a 45° angle with an initial speed of 20 m/s, it's essential to use the equations of motion. The horizontal distance (x) can be calculated using x = Vox * t, while the vertical distance (y) is determined by y = Voy * t - 1/2 * g * t^2. The jumper's initial velocity components need to be resolved based on the angle of launch and the slope. The challenge lies in the unknown height of the slope, which complicates the calculations. A clear understanding of projectile motion principles is crucial for solving this problem effectively.
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Homework Statement



I have to calculate disctance that body(for example a jumper) that jumps of a slope at 45 degrees with initial speed of 20 m/s and under 15 degrees.

Homework Equations


x=Vox * t
y=Voy*t - 1/2*g*t*t

y=ax + b, b=0 => y=ax
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The Attempt at a Solution



I don't know where to start. I can solve a simillar example where body has been shot from a building with a certain initial speed and under under 30°. It is simillar to this picture, but here it is difference that we shoot from a slope and we don't know at what height we only know that slope is placed under 45° with a surface.

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