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2D Projectile Motion

  • Thread starter MattPalmer
  • Start date
  • #1
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
 

Answers and Replies

  • #2
3,740
417
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]
 
  • #3
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?
 
  • #4
CAF123
Gold Member
2,902
88
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.
 
  • #5
3,740
417
Would that mean that I need to solve the equation using the quadratic formula?
Yes.
 
  • #6
rl.bhat
Homework Helper
4,433
7
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.
 
  • #7
Got it, thanks guys
 

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