# Projectile Motion launch speed

1. Jan 27, 2013

### aaronfue

1. The problem statement, all variables and given/known data

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 vA necessary to make the jump.

Image attached.

2. Relevant equations

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

xB = xo + (vo)xt + $\frac{1}{2}$at2

yB = yo + (vo)yt + $\frac{1}{2}$at2

3. 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): vAsin30 - $\frac{9.81}{2}$t2 = 0

Equation 2 (horizontal motion): vAcos30t = 25

Is this the correct approach? Are there any errors?

#### Attached Files:

• ###### Motorcycle jump.JPG
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2. Jan 28, 2013

### Staff: Mentor

That is a very good choice.

Correct, and this will allow you to get va.

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.

Last edited by a moderator: Jan 29, 2013
3. Jan 29, 2013

### jing2178

Equation 1 t missing

vAsin(30)t - $\frac{9.81}{2}$t2 = 0

Equation 2 to stop confusion with cos(30t)

vAcos(30)t = 25