An interesting twist!
Let the initial horizontal velocity be vx, and the initial vertical velocity by vy (both in ft/s).
Time taken to travel a horizontal distance of D=59 yards = 177 ft is
D/vx
Time t taken to go up and fall back to ground level, based on the acceleration due to gravity g=32.2 ft/s² is:
uy*t-(1/2)gt²=0
t(uy-gt/2)=0
where t=0 is a solution (initial cond.)
So t=2*uy/g is the time it takes to land on the ground.
Equate with time for horizontal distance, we get
2*uy/g = D/vx
ux*uy = gD/2 = 32.2*177/2 = 2849.7 ft²/s²
For minimal effort, we can set
ux=uy=√(2849.7)= 53.3826 ft/s
Since ux=uy, angle with horizontal is 45°.
Note that the angle can vary, depending on the value of one of the two quantities, vx or vy.
So i need to figure out what the angle and speed was of a foot ball that traveled 59 yards how can i figure that out? what formula can i use?
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