Let's assume 20 degrees upward angle for launch
The velocity vector can be separated into a horizontal component and a vertical component.
So vector Vtotal = vector Vvertical + vector Vhorizontal
By pythagorean theorem we get:
Vvertical = Vtotal * sin(20)
Vhorizontal = Vtotal * cos(20)
Vertical distance will net to zero, and is goverened by the equation distance = Velocity * time + 1/2 * acceleration * time^2. Acceleration is -9.8 meters/second^2. (Negative because intitial velocity is up
and gravity is pulling down
0 = Vvertical * hangtime + 1/2 (-9.8 meters/second^2) * hangtime^2
0 = Vtotal * sin(20) * hangtime - 4.9 meters/second^2 * hangtime^2
Divide by hangtime
0 = Vtotal * sin(20) - 4.9 meters/second^2 * hangtime
Vtotal = (4.9 meters/second^2 * hangtime)/sin(20)
Next, compute horizontal component:
Distancehorizontal = Vhorizontal * hangtime
(I'm gonna cheat a little and say 10 feet = 3 meters when it really equals 3.048 meters)
3 meters = Vtotal * cos(20) * hangtime
Vtotal = 3 meters /(cos(20) * hangtime)
Substitute Vtotal above
(4.9 meters/second^2 * hangtime)/sin(20) = 3 meters /(cos(20) * hangtime)
Solve for hangtime
hangtime = squareroot(3 meters * tan(20)/4.9 meters/second^2)
Which is 0.47205817325674093410684661156673 seconds.
This means that Vhorizontal = 6.355 meters/sec
Vhorizontal = 6.355 meters/sec = Vtotal * cos(20)
Vtotal = 6.763 meters/sec
Vvertical = Vtotal * sin(20) = 2.313 meters/sec
Now, how much time will an additional 5 feet drop add to the hangtime?
Fortunately, there's a handy calculator http://hyperphysics.phy-astr.gsu.edu/Hbase/mot.html
(scroll down to "motion example).
add'l hangtime = 0.36550106447852343 seconds
Additional distance covered = 2.32275926476101639765 meters = 7.62060126 feet.