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A ball is struck such that its max height is equal to its range. determine the angle its struck at. Air resistance is ignored and the ground is horizontal.

my reasoning is that:

using vectors Xsin(angle) = Xcos(angle) so tan(angle) = x/x so inverse tan of 1 is 45degrees

Would this be the right approach to the question Id ask my lecturer but im at home Cheers

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closed as off-topic by tpg2114, Qmechanic Nov 3 '13 at 2:28

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – tpg2114, Qmechanic
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I'm adding the "homework" tag because, even though this may not be a homework assignment specifically, it has an educational purpose so it's effectively similar. –  David Z May 5 '11 at 19:29

1 Answer 1

up vote 3 down vote accepted

Range $ = v^2 \sin(2\theta) / g = (v \sin \theta)^2 / 2g = $ Max height

$4 \sin\theta \cos\theta = \sin\theta \sin\theta $ because $\sin2\theta = 2 \sin\theta \cos\theta$

$ 4 \cos \theta = \sin\theta $

$ \tan\theta = 4$

$\theta = \tan^{-1} 4$

Problem is $X\cos\theta = X\sin\theta$ equates horizontal and vertical components of $X$ (whatever $X$ is), which isn't same as equating range of projectile to maximum possible height when thrown at certain angle.

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Is $\vec{X}cos\theta$ right? or should I use $Xcos\theta$? –  Pratik Deoghare May 5 '11 at 20:15
2  
You should say $X\cos\theta$. There's no arrow over the $X$, because that symbol here refers just to the magnitude of the vector $\vec X$, not to the entire vectorial quantity. Also, the functions $\sin,\cos$ are traditionally written in roman, not italic type. In TeX syntax, putting a backslash in front of the names sin and cos takes care of that. –  Ted Bunn May 5 '11 at 20:43

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