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2D Projectile Motion

 
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Sep9-12, 05:25 PM   #1
 

2D Projectile Motion


A ball is tossed from an upper story window of a building. It has an initial velocity of 8.00 m/s at an angle of 20 degrees below the horizontal. How long does it take the ball to reach a point 10.0m below the level of launching.

I tried using the equation
yf=vi(sin(theta))-.5gt^2
Which I simplified to
(yf-vi(sin(theta)))/.5g = t^2
Then
(10-8(sin(20)))/(4.91) =t^2
Finally t=1.23, however my book says I should be getting 1.18s, what am I doing incorrectly?
Thanks in advance
 
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Sep9-12, 07:54 PM   #2
 
Your equation is missing a "t" in the term containing the initial velocity.
The general form should be
[tex]y=y_0+v_{y0}t+1/2gt^2[/tex]
 
Sep9-12, 09:01 PM   #3
 
Quote by nasu View Post
Your equation is missing a "t" in the term containing the initial velocity.
The general form should be
[tex]y=y_0+v_{y0}t+1/2gt^2[/tex]
Would that mean that I need to solve the equation using the quadratic formula?
 
Sep10-12, 03:19 AM   #4
 

2D Projectile Motion


You are missing a 't' after the [itex] v_o\sin\theta [/itex] term. After you have this, rearrange your equation into the form [itex] at^2 + bt +c = 0 [/itex] and use methods of solving quadratics.
 
Sep10-12, 08:03 AM   #5
 
Quote by MattPalmer View Post
Would that mean that I need to solve the equation using the quadratic formula?
Yes.
 
Sep10-12, 09:50 AM   #6
 
Recognitions:
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Check the sign convention. If g is negative in the downward direction then yf -yo and vi*sin(theta) should be negative in the downward direction.
 
Sep10-12, 12:23 PM   #7
 
Got it, thanks guys
 
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