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Part c is basically asking for the trajectory, right?

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Part c is basically asking for the trajectory, right?

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Doc Al

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In a sense, yes, but only one aspect of it. How would you go about finding the maximum height?frankfjf said:Part c is basically asking for the trajectory, right?

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To be honest I'm not sure.

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Doc Al

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I think that the formula involved would be the vertical motion equation for velocity, Vy = Vi*sin(angle) - gt.

Well actually, if I took half of the time, wouldn't it be at its highest point then?

- #6

Doc Al

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Think about it. Imagine throwing a ball straight up in the air. Initially it moves up (velocity = +); after reaching the highest point, it starts moving down (velocity = -). So what can you say about the speed at the highest point?frankfjf said:That's just it, I have no idea what to plug in for determining the velocity at the highest point.

Good.I think that the formula involved would be the vertical motion equation for velocity, Vy = Vi*sin(angle) - gt.

Only if the initial and final positions were at the same height, which is not true in this problem.Well actually, if I took half of the time, wouldn't it be at its highest point then?

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Well, at the highest point the speed would be zero, since it's going to start falling immediately after hitting the highest point.Doc Al said:Think about it. Imagine throwing a ball straight up in the air. Initially it moves up (velocity = +); after reaching the highest point, it starts moving down (velocity = -). So what can you say about the speed at the highest point?

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Doc Al

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Exactly right. Now make use of that fact.

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If the formula Vy = Vi*sin(angle) - gt is used, does the zero replace Vy or Vi?

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Doc Al

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What do you think? (Does the *initial *velocity change?)

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No, but if I put in zero for Vy, I'm left without any variables.

- #12

Doc Al

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You can solve for the time; that's what you need.

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Ahhh I got it now, thanks!

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