- #1
jahrollins
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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.
