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August 20, 2014

August 20, 2014

Posted by **Kate** on Monday, August 8, 2011 at 5:38pm.

- math -
**MathMate**, Monday, August 8, 2011 at 6:12pmAn 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.

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