Quote:
Originally Posted by Jeff Lebowski
No, you can solve for it.
He threw the ball 16 ft above his head. I assumed he is six feet tall, so it went a total of 22 ft above the ground.
In case you are interested, here is how I did it:
The ball is going to start with some unknown velocity (v), go up in the air in a parabolic arc and then come back to the ground. As it passes the point from which he threw it (6 ft) on the way down it will be traveling with the same velocity (v) it started with but in the opposite direction. There are two parts to the path:
part a (up and down to his head):
height = ya
time = ta
part b (from his head to the ground):
height = yb = 6
time = tb
Now we can set up some equations:
ya = v^2/(2*g)
(g= gravitational constant = 32.2 ft/sec^2)
yb = 6 = v*tb + (1/2)*g*tb^2
ta + tb = 2.36
ta = (2*v)/g
I threw these in a spreadsheet and solved for v using the Goal Seek function. Final values:
v: 31.9 ft/s
ta: 1.98 s
tb: 0.38 s
ya: 15.8 ft
Took me about ten minutes to solve. Yes, I am a geek and proud of it.
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Big guy, now show us the PAT would have been blocked regardless.
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