- #1
Saladsamurai
- 3,020
- 7
[SOLVED] Projectile motion (am I on drugs?)
I can't believe how much time I have spent on this.
Constant acceleration
Here is the data I am using to solve for
[itex]V_a[/itex]
[itex]x_0=0[/itex],
[itex]x_f=30[/itex],
[itex]y_0=7[/itex],
[itex]y_f=10[/itex],
[itex]a_y=-g=-32.2[/itex],
[itex]V_{0y}=V_a\sin30[/itex],
[itex]V_{0x}=V_a\cos30[/itex]
From x direction:
[tex]\Delta X=V_{0x}t\Rightarrow t=\frac{30}{V_a\cos30}[/tex]
From y direction:
[tex]\Delta Y=V_{oy}+.5a_yt^2[/tex]
[tex]\Rightarrow 3= V_a\sin30*(\frac{30}{V_a\cos30})-16.1[\frac{30}{V_a\cos30}]^2[/tex]
[tex]\Rightarrow 3=30\tan30-16.1*(\frac{40}{V_a})^2[/tex]
Which gets me a negative under the radical!
Homework Statement
I can't believe how much time I have spent on this.
Homework Equations
Constant acceleration
The Attempt at a Solution
Here is the data I am using to solve for
[itex]V_a[/itex]
[itex]x_0=0[/itex],
[itex]x_f=30[/itex],
[itex]y_0=7[/itex],
[itex]y_f=10[/itex],
[itex]a_y=-g=-32.2[/itex],
[itex]V_{0y}=V_a\sin30[/itex],
[itex]V_{0x}=V_a\cos30[/itex]
From x direction:
[tex]\Delta X=V_{0x}t\Rightarrow t=\frac{30}{V_a\cos30}[/tex]
From y direction:
[tex]\Delta Y=V_{oy}+.5a_yt^2[/tex]
[tex]\Rightarrow 3= V_a\sin30*(\frac{30}{V_a\cos30})-16.1[\frac{30}{V_a\cos30}]^2[/tex]
[tex]\Rightarrow 3=30\tan30-16.1*(\frac{40}{V_a})^2[/tex]
Which gets me a negative under the radical!