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I know the angle at which a projectile is launched, how far it needs to go, and also the maximum height. How can I find the initial velocity needed (disregarding air resistance)?

Currently, I am using:

$v_f^2=v_y^2-2g\Delta y$

From that, I can find the velocity in the $y$ direction needed for the projectile to reach the known maximum height.

Next, I am finding the time it takes to reach that maximum height by using the found y velocity.

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There are loads of articles out in Googleland on calculating the trajectory of a projectile, and indeed searching this site for "projectile" finds lots of related questions. – John Rennie Sep 18 '12 at 11:30
    
    
why did you write the kinematic equation with negative sign there? – Nasser Oct 11 '13 at 2:15
    
seriously, you have to do some work on your own before asking questions to the esteemed contributors here. Further, from experience, I suggest that you derive the equations for projectile motion from the start, i.e., by considering projectile motion as being made of two linear motions. – don_Gunner94 yesterday

If you know the Maximum height and Range then just use $y = u_yt-\frac12gt^2 $and $x=u_xt$
Eliminate t and get the value of u directly . This is also the basic concept behind the well known equation of trajectory : $y=xtan\alpha-\frac {gx^2} {2u^2cos^2\alpha}. $If you have a good memory you may also use this directly.

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HINT: Use $H_{max}=\frac{u^2sin^2\theta}{2g}$ (max height) and $R=\frac{u^2sin2\theta}g$ (Range).

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protected by Qmechanic yesterday

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