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<p>Ok, for our lab on projectile (2-d) motion,</p>
<p>We know : </p>
<p>Angle of launcher(little canon) with respect to surface
height of launcher abover surface
average horizontal distance ball traveled at that angle/height
-9.8 gravity force </p>
<p>and are required to find the launch speed. What equation should I use to accomplish that?</p>
<p>Thank you.</p>
<p>You'd actually need a combination of equations relating the vertical motion and the horizontal motion without using time.</p>
<p>For horizontal distance
vx=dx(t) vx is the horizontal component of velocity, given by vcosθ, and dx is the horizontal distance.</p>
<p>dy=vyt+4.9y^2, where vy=vsinθ</p>
<p>Any more info given?</p>
<p>There is a mention of the range/trajectory equation since we don't have any value for time. But i thought this was only used when the place where the projectile landed was equal to that of the release point. In this case, it lands on the floor from a table, can I still use that equation ?</p>
<p>See Rod Nave's Hyperphysics page at <a href="http://hyperphysics.phy-astr.gsu.edu/hbase/hph.html%5B/url%5D">http://hyperphysics.phy-astr.gsu.edu/hbase/hph.html</a></p>
<p>Range can be described as a function of launch angle and intitial velocity.</p>
<p>The ballistics page has nice diagrams, derivations and Java calculators.
<a href="http://hyperphysics.phy-astr.gsu.edu/hbase/traj.html#tra4%5B/url%5D">http://hyperphysics.phy-astr.gsu.edu/hbase/traj.html#tra4</a></p>
<p>First use the angle of launch and the height above the surface to figure out the initial x and y velocities by doing for example.... 25sin12 where 12 is the angle... this would normally give the y velocity and do cos for x but make sure its not the other way around. Once you get the velocities simply use the equation v^2= v(i)^2 + 2ad and find it that way.</p>