Projectile Motion - given angle, time; need displacement

[jahrollins]

Homework Statement

A ball is thrown at 60 degrees above x-axis. It remains in the air for 2.9 seconds. What is the final displacement in the x (range) if the ball was thrown with the same initial velocity, but at an angle of 45 degrees above x-axis instead?

Homework Equations

y = (Voy * t) + (1/2 * ay * t^2)
Voy = Vo * sin(theta)
Vox = Vo * cos(theta)
x = 1/2(Vox + Vx)t

The Attempt at a Solution

So plugging into the first equation I get:
y = (Voy * t) + (1/2 * ay * t^2)
0 = (Voy * 2.9) + (1/2 * -9.8 * 2.9^2)
Voy = 14.21

Taking the result into the second equation I get the initial velocity:
Voy = Vo * sin(theta)
14.21 = Vo * sin(60)
Vo = 16.41

Am I on the right track so far? From here I would think I'd take it into the first equation again with the 45 degree angle so:
y = (Voy * t) + (1/2 * ay * t^2)
0 = (16.41sin(45)*t) + (1/2 * -9.8 * t^2)
0=11.60t-4.9t^2
t = 2.37
So I've got the time, and Vox from Vox = Vo * cos(theta) since Vo is the same for both throws.

For the equation
x = 1/2(Vox + Vx)t
(substitute Vox = Vo * cos(theta))
x = 1/2(16.41cos(45) + Vx)2.37

All I'm missing is Vx, I am lost where to go from here.

 

[ehild]

Is there any horizontal force component?

ehild

 

[jahrollins]

Is there any horizontal force component?

ehild

No...

 

[ehild]

No force, no acceleration. What does it mean for vx?

ehild

 

[jahrollins]

Is it Vx = Vo?

Vx = Vo + 1/2*ax*t
Vx = Vo + 1/2*0*t
Vx = Vo

 

[ehild]

Yes, you have got it !

ehild

 

[jahrollins]

I think I understand now, thanks a bunch. :>