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- Thread starter BananaRed
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- #1

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- #2

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Need to be more specific. What's the landscape look like?

cookiemonster

cookiemonster

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that's all I was given

- #4

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well first thing, I'm guessing your instructor wants you to come with an equation.

But first things first. When you release the ball, the point in the trajectory where it reaches the level it lands on, the ball will have to be traveling at 45 degrees. So at the level of the landing point, your horizontal and vertical velocities must be equal. It's easy to visualize when throwing at a level higher than your initial, but harder when it is lower.

For the solution, think position equation, and vectors.

So first we have [tex] \sin(\theta) v - \cos(\theta) v = a t [/tex]

Solving t for position equation ( [tex] s(t) = \frac{1}{2}at^2 + vt + s [/tex] ) using the quadratic equation, 1/2 acceleration cancels and we're left with

[tex] sin(\theta) v - \cos(\theta) v = \frac{-v \pm \sqrt{v^2 - 4 \frac{a}{2} s}}{1} [/tex]

I'll let you do the rest, or else it won't be your work.

And someone plz correct me if I'm wrong.

But first things first. When you release the ball, the point in the trajectory where it reaches the level it lands on, the ball will have to be traveling at 45 degrees. So at the level of the landing point, your horizontal and vertical velocities must be equal. It's easy to visualize when throwing at a level higher than your initial, but harder when it is lower.

For the solution, think position equation, and vectors.

So first we have [tex] \sin(\theta) v - \cos(\theta) v = a t [/tex]

Solving t for position equation ( [tex] s(t) = \frac{1}{2}at^2 + vt + s [/tex] ) using the quadratic equation, 1/2 acceleration cancels and we're left with

[tex] sin(\theta) v - \cos(\theta) v = \frac{-v \pm \sqrt{v^2 - 4 \frac{a}{2} s}}{1} [/tex]

I'll let you do the rest, or else it won't be your work.

And someone plz correct me if I'm wrong.

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