# Homework Help: Help with kinematics

1. Sep 17, 2007

### devilsangels287

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

A sailboard is sailing at 6.5 when a gust of wind hits, causing it to accelerate at 0.48 at a 35 angle to its original direction of motion.

If the acceleration lasts 6.3 , what is the board's net displacement during the wind gust?

2. Relevant equations

Ok so far, I have drawn all of the motion graphs.
So I got

X Y

X_0= 0m Y_0= 0m
X_F=? Y_F=?
V_0x= 6.5m/s v_oy=0m/s
V_fy= 6.5m/s v_fy=?
a= 0 m/s^2 a= .48m/s^2
t= 6.3secs t=6.3secs

3. The attempt at a solution

And assuming those are correct. I got v_fy= 4.55 m/s by using 6.5tan35 = v_fy

then I used this equation: x=x_0+v_ot+(1/2)at^2 and got

X_f= 40.95m
Y_f= 9.5m
And then I added them together and then its 50.45m??? Is this correct, because online it said it was wrong?

Thanks for helping!

2. Sep 17, 2007

### learningphysics

suppose it's initially moving eastbound at 6.5m/s. ie: northbound velocity = 0.

so acceleration 0.48m/s^2 east 30 degrees north.

What is the component of acceleration in the east/west direction?

What is the component of acceleration in the north/south direction?

So you can divide the problem into the east/west part (to get the displacement east/west), and the north/south part (to get the displacement north/south)... and work them separetly... each part is a uniform acceleration problem.

3. Mar 7, 2010

Need help with kinematics

I figured out this problem:

You step off a cliff 30 meters high.
A. How long will it take to hit the water below?
0=30-4.9t^2
4.9t^2=30
t^2= 6.122
t= 2.47

B. What is your velocity (mph) when you hit the water?
Vty= Voy-9.8t
Vty-Voy-9.8(2.47)
Vty=-24.25 m/s = 54.223 mph

Can anyone help me solve this one, based on the above problem?:

2. How high would the cliff have to be in 1. above if your velocity hitting the water was 100 mph?

Thanks!