Calculating Projectile Motion: Time to Reach 10m Below Launch Point

[MattPalmer]

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

 

[nasu]

Your equation is missing a "t" in the term containing the initial velocity.
The general form should be
$$y=y_0+v_{y0}t+1/2gt^2$$

 

[MattPalmer]

Your equation is missing a "t" in the term containing the initial velocity.
The general form should be
$$y=y_0+v_{y0}t+1/2gt^2$$

Would that mean that I need to solve the equation using the quadratic formula?

 

[CAF123]

You are missing a 't' after the $v_o\sin\theta$ term. After you have this, rearrange your equation into the form $at^2 + bt +c = 0$ and use methods of solving quadratics.

 

[nasu]

Would that mean that I need to solve the equation using the quadratic formula?

Yes.

 

[rl.bhat]

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.

 

[MattPalmer]

Got it, thanks guys