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2D Projectile Motion |
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| Sep9-12, 05:25 PM | #1 |
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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 |
| Sep9-12, 07:54 PM | #2 |
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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 |
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| Sep10-12, 03:19 AM | #4 |
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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.
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| Sep10-12, 08:03 AM | #5 |
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| Sep10-12, 09:50 AM | #6 |
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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.
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| Sep10-12, 12:23 PM | #7 |
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Got it, thanks guys
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