Physics
posted by George .
Can someone help me get started on this problem?
A launcher fires a projectile with a velocity of 6.31m/s. Determine the horizontal distance of the projectile (in the air) at 5 degrees. Recalculate this distance at 10, 15, 20.... (5 degree increments) up to 90 degrees (at which point the horizontal distance will be zero). Graph this data (Distance vs. Launch Angle). Distance will be on the Yaxis and launch angle will be on the Xaxis.

The equation for range is
Distance = 2(V^2/g)sinX*cosX
= V^2/g sin(2X)
where X is the launch angle. In your case V^2/g is 4.05 m. The projectile won't go very far. 6.31 m/s is the speed of a slow runner.
You curve will be the half of a "sine wave.", with zeros at X=0 and X = 90 degrees and a maximum at X=45 degrees 
It is the initial vertical velocity that determines how long it is in the air, and that determines how far it will go.
time in air:
initialverticalvelocity viv=6.31*sinTheta
hf=hi+viv*t4.9t^2. Hf=hi=0, solve for timeinair t.
t=viv/4.9=6.31sinTheta/4.9
horizontal distance:
horizontal velocity vih=6.31cosTheta
distance=6.31cosTheta* t=
= 6.31^2 sinTheta*CosTheta /4.9
check all that.