Parallel boat to dock jump. Sense of Humor required

1. Mar 3, 2013

burton95

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

The deck of the boat is parallel to the dock as you jump from the boat and the direction of your velocity as you jump is 30° above the horizontal. If the boat was stationary when you jumped and 1.5 m away from the dock, what is the horizontal speed of the boat (in m/s) after you jump to the dock? Assume that your mass is 70 kg and the mass of the boat (including the captain) is 1400 kg.

2. Relevant equations

Pf = Pi

(Fnet) = ma

3. The attempt at a solution

0 = m1vf1 + m2vf2

vf2 = -m1vf1/m2

(fnet)x = Fcosθ = m1a1
a= Fcosθ/m1

(fnet)y = -Fsinθ + n - m1g = 0

2. Mar 3, 2013

haruspex

How you tackle this depends on how precise you want the answer. Strictly speaking, you should allow for the fact that the boat's movement will rob you of some horizontal speed. It then becomes important whether the take-off angle is relative to the boat as a frame of reference or relative to the dock. Let's assume the second.
Suppose your horizontal velocity relative to the dock is u. What was your vertical take off velocity? How long will it take to reach the dock? What height will you be at then? What does that tell you about u?

3. Mar 3, 2013

burton95

I took off the long funny part. Kind of inappropriate

burton

4. Mar 3, 2013

burton95

relative to the dock I'm jumping 60 degrees. u = F*Sinθ and v = Fcosθ right?

I don't understand how I get the velcotiy of my jump

5. Mar 3, 2013

haruspex

You may have misunderstood my point. If you jump at some angle to the horizontal relative to stationary ground, the angle of your jump will appear different to a passer by moving at some speed relative to the ground. Let's agree the question means 30 degrees to the horizontal as judged by a person standing on the dock.
That depends what θ, F, u and v mean. Please define.
It needs to be just sufficient to reach the dock.