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November 28, 2014

November 28, 2014

Posted by **Vinny** on Thursday, November 14, 2013 at 8:45pm.

- Physics -
**rosa**, Thursday, November 14, 2013 at 9:40pmIn this following example, just keep v_0=9.8v_x and solve in the same way...

The speed of the projectile can be broken into two velocity components. Vy and Vx. At maximum height Vy is equal to 0. The velocity vector V=(Vy^2+Vx^2)^1/2

We also know, that the initial speed of the projectile is 8 times the amount of max height.

The initial velocity of the projectile is Vo, and that can be broken into two components as well. Vox and Voy. From vectors, Vo=(Vox^2+Voy^2)^1/2

Vo=8 Vx.... Substitute we get... Vox^2 + Voy^2 = 64*Vx^2.

Tan(x)=Voy/Vox..... Voy= Tan(x)*Vox.

Substitute again.

Vox^2 + (Tan(x)*Vox)^2 = 64*Vx^2.

Now one final equation. Vx=Vox because gravity does not act in the X direction but only the why.

So we divide by Vox^2 and we get

1+ (Tan(x))^2=64

then solve for Tan (x) and x= arctan(...)

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