%3Cp%3ELauren%20is%20jumping%20across%20a%2010.0m%20deep%20ditch%20on%20her%20mo
ID: 3894033 • Letter: #
Question
%3Cp%3ELauren%20is%20jumping%20across%20a%2010.0m%20deep%20ditch%20on%20her%20motorcycle.%0AShe%20takes%20off%20from%20a%2010.0m%20high%20ramp%20that%20has%20an%20angle%20of%2030%0Adegrees%20relative%20to%20the%20ground%20and%20hopes%20to%20cross%20the%20200.0m%20wide%0Aditch.%20%20What%20is%20the%20minimum%20speed%20she%20needs%20to%20land%20on%20the%0Aother%20side%3F%20Ignore%20air%20resistance.%3C%2Fp%3E%0A%3Cp%3E%3Cbr%20%2F%3E%3C%2Fp%3E%0A%3Cp%3EChoose%20x%20to%20the%20right%20and%20y%3Dup%3C%2Fp%3E%0A%3Cp%3EChoose%20initial%20position%20when%20the%20motocycle%20leaves%20the%20ramp%20as%0AXo%3D0m%2C%20Yo%3D10m%3C%2Fp%3E%0A%3Cp%3EWhen%20the%20motocycle%20lands%2C%20x%3D200m%2C%20and%20y%3D0m%3C%2Fp%3E%0A%3Cp%3E%3Cbr%20%2F%3E%3C%2Fp%3E%0A%3Cp%3EThe%20equations%20or%20motion%20are%3C%2Fp%3E%0A%3Cp%3Ey(t)%3D10%2B(Vosin30)t-4.9t%5E2%3C%2Fp%3E%0A%3Cp%3Ex(t)%3D(vocos30)t%3C%2Fp%3E%0A%3Cp%3E%3Cbr%20%2F%3E%3C%2Fp%3E%0A%3Cp%3EChoose%20the%20time%20t%20for%20when%20the%20motocycle%20lands%3C%2Fp%3E%0A%3Cp%3Ey%20equation%3A%200%3D10%2B(vosin30)t-4.9t%5E2%3C%2Fp%3E%0A%3Cp%3Ex%20equation%3A%20200%3D%20(vocos30)t%3C%2Fp%3E%0A%3Cp%3E%3Cbr%20%2F%3E%3C%2Fp%3E%0A%3Cp%3ESubstitute%20t%3D200%2F(vocos30)%20into%20y%20equation%2C%20solve%20for%20Vo%3C%2Fp%3E%0A%3Cp%3E%3Cbr%20%2F%3E%3C%2Fp%3E%0A%3Cp%3EI%20know%20the%20answer%20is%2045.6%20m%2Fs%20but%20I%20have%20no%20idea%20how%20to%20solve%0Afor%20it.%20%26nbsp%3BPlease%20help.%20%26nbsp%3BReally%20appreciate%20it.%3C%2Fp%3E%0AExplanation / Answer
better yet, let vo = 200/(cos 30 t) giving you
0 = 10 + (vo sin 30)(200 / cos 30 t) * t - 4.9 t^2
0 = 10 + 200 tan 30 - 4.9 t^2
Solve for t, then plug back into the equation for vo and you will get 45.6.
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