Solve Kinematics Homework: Wind Gust Displacement of Sailboard

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A sailboard experiences a gust of wind that accelerates it at 0.48 m/s² at a 35-degree angle for 6.3 seconds, prompting a discussion on calculating net displacement. Initial calculations yielded a final velocity in the y-direction of 4.55 m/s and a total displacement of 50.45 m, which was questioned for accuracy. The discussion emphasizes breaking down the problem into east/west and north/south components to solve for displacement separately. Additionally, a related problem involving free fall from a 30-meter cliff was solved, determining the time to hit the water and the impact velocity. The conversation highlights the importance of correctly applying kinematic equations to solve motion problems.
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Homework Statement



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?


Homework 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

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!
 
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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.
 
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!
 
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