What Launch Speed Does the Motorcyclist Need to Clear the Jump?

[aaronfue]

Homework Statement

The motorcyclist attempts to jump over a series of cars and trucks and lands smoothly on the other ramp, i.e., such that his velocity is tangent to the ramp at B. Determine the launch speed v_A necessary to make the jump.

Image attached.

Homework Equations

a = -9.81 \frac{m}{s^2}

x_B = x_o + (v_o)_xt + \frac{1}{2}at²

y_B = y_o + (v_o)_yt + \frac{1}{2}at²

The Attempt at a Solution

I set my point of origin for my coordinate system right at point A (end of ramp). I wanted to keep from using the 3 m height of both ramps. (Would I be able to solve this problem this way?)

I came up with the following equations:

Equation 1 (vertical motion): v_Asin30 - \frac{9.81}{2}t² = 0

Equation 2 (horizontal motion): v_Acos30t = 25

Is this the correct approach? Are there any errors?

 

[mfb]

I set my point of origin for my coordinate system right at point A (end of ramp). I wanted to keep from using the 3 m height of both ramps. (Would I be able to solve this problem this way?)

That is a very good choice.

Equation 1 (vertical motion): v_Asin30 - \frac{9.81}{2}t² = 0

Equation 2 (horizontal motion): v_Acos30t = 25

Correct, and this will allow you to get v_a.

such that his velocity is tangent to the ramp at B

This follows directly from the requirement that the motorcyclist hits the ramp and the geometry of the problem, I don't know why this was added here. You could add a comment here in some way, or even check the condition.

 

[jing2178]

I came up with the following equations:

Equation 1 (vertical motion): v_Asin30 - \frac{9.81}{2}t² = 0

Equation 2 (horizontal motion): v_Acos30t = 25

Is this the correct approach? Are there any errors?

Equation 1 t missing

v_Asin(30)t - \frac{9.81}{2}t² = 0

Equation 2 to stop confusion with cos(30t)

v_Acos(30)t = 25