# Search results

1. ### Projectile Motion on Inclined Surface Question

Hey TSny! Thank you it was a sign error! xi = -61.07 yi = -45.80 Range = sqrt[ (-61.07)^2 + (-45.80)^2 ] = sqrt(5827.17) = 76.33. close enough Thank you both for all your help. I was really confused.
2. ### Projectile Motion on Inclined Surface Question

Hi TSny, thanks for the quick reply. - the 'd' in the cosd does denote that the argument is in degrees (from matlab, dunno why I included it here) - the 'i' does mean intersection. apologies. - I will look into the sign error, hopefully that fixes it.
3. ### Projectile Motion on Inclined Surface Question

Hi mfb, thank you for the quick response. - Oh! Yes my my coordinate system is a right handed one: x positive right, y positive up - Unfortunately, my final answer is different than the solution. ANS: Range = 76.5 meters Range = sqrt( xi^2 + yi^2) = sqrt( 17.126^2 + 12.8449^2) = 21.4 meters...
4. ### Projectile Motion on Inclined Surface Question

Homework Statement Water is sprayed at an angle of 90° from the slope at 20m/s. Determine the range R. PLEASE SEE ATTACHMENT Homework Equations [/B] Kinematic Equations: acceleration: A = 0i +g(-j) + 0k; velocity: dV/dt = A; position: dR/dt = V; Origin set at the point...