# Projectile Motion

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.

• 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. Commented Sep 18, 2012 at 11:30
• Commented Sep 18, 2012 at 11:32
• why did you write the kinematic equation with negative sign there? Commented Oct 11, 2013 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. Commented Apr 30, 2016 at 14:49
• Use the range equation. en.wikipedia.org/wiki/Range_of_a_projectile Commented Oct 14, 2019 at 1:10

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.

$$v_0=\sqrt{\frac{gL}{2\cos^2\theta_0\left(\tan\theta_0-\frac{h}{L}\right)}}$$

where $h =$ Height and $L =$ Lenght.

Hope you find it useful.

• would you please elaborate? Commented Apr 30, 2016 at 15:14
• I don't know how to use notation, this is barely my second post, so I will try to explain as accurately as I can. By the Newton's Laws, we got that if we know the V0 and the Angle of the projectile, then we could accurately detail where will be the position in the coordenates P(X;Y), from a point O(0;0), by a time x: x=v0.t.cos(thita) y=v0.t.sen(thita)-1/2g.t^2 And the components of the Vx, and Vy, in any instant t, are respectly: Vx=V0.cos(thita) Vy=V0.sen(thita)-g.t Now you isolate the variable t, and insert it on the initial equation to end up on the first i posted. Check that -sen/cos=tg Commented Apr 30, 2016 at 17:15
• I didn't realise you were new, sorry! Welcome to SE :-) If you want to learn how to typeset math formulas, I suggest you to read this basic tutorial. I'll edit your post to input the math. When I'm done, click on the edit button to see how I did it. Have fun, and thank you for your contribution! Commented Apr 30, 2016 at 17:21

HINT: Use $$H_{\max}=\dfrac{u^2\sin^2\theta}{2g}$$ (max height)

and $$R=\dfrac{u^2\sin2\theta}g$$ (range).