Yahoo Answers: Answers and Comments for Physics help...Motion in 2 dimensions...(Uniform Circular Motion/Projectile Motion) Combination Problems? [Physics]
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From RF
enUS
Sat, 09 Jan 2010 21:10:24 +0000
3
Yahoo Answers: Answers and Comments for Physics help...Motion in 2 dimensions...(Uniform Circular Motion/Projectile Motion) Combination Problems? [Physics]
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https://answers.yahoo.com/question/index?qid=20100109211024AAoajAM
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From Physicsquest: First of all lets find tip speed (tangential v...
https://answers.yahoo.com/question/index?qid=20100109211024AAoajAM
https://answers.yahoo.com/question/index?qid=20100109211024AAoajAM
Sun, 10 Jan 2010 02:38:02 +0000
First of all lets find tip speed (tangential velocity)
3.7 Hz = 3.7 cycles per second (cps)
3.7 cps x (2 x pi) x 0.52 = 12.09 m/s (tangential velocity)
12.09sin(38) = initial vertical velocity = 7.4426 m/s
lets find max height:
v^2 = zero at highest point = u^2  2gh
7.4426^2 / (2g) = h = 2.826 m
2.826 + 1.2 = 4.026 m ( max h)
The horizontal velocity = 12.09cos(38) = 9.527 m/s
Time to max h = (4.0261.2) / 4.9 = t^2 = 0.5767, sqrt = t = 0.759 secs
Time for rock to fall to reach 2.4 m = 4.026  2.4 = 1.626 m
1.626 m = 1/2at^2 since the rock falls from rest at max h
1.626/4.9 = t^2 = 0.3318, sqrt = t = 0.576 secs
max time = 0.576 + 0.759 = 1.335 secs
1.335 x horizontal velocity ( 9.527) = 12.72 m
The above is what i think the answer is.
Now lets look at the answer you arrived at?
You got 1.61 m
Ok, i'll work it out:
v^2 = u^2  2gh
u^2 vertical = 7.4426 m/s
7.4426^2  2gh = v^2
2gh = 2 x 9.8 x ( 2.4  1.2 ) = 23.52
55.39  23.52 = v^2 = 31.87, sqrt = 5.645 m/s (vertical velocity at impact )
t = (2.4  1.2) / ((7.4426 + 5.645)/2) = 0.1833 secs
0.1833 x horizontal velocity (9.527) = 1.746 m
I can't see any one in the right mind doing this at such short distance? so i think the answer will be 12.72 m.
I hope this was done step by step for you to understand.

From MS: V(0)=rw ,w=2*pi*f, w=23.248,v(0)=0.52*23.248=1...
https://answers.yahoo.com/question/index?qid=20100109211024AAoajAM
https://answers.yahoo.com/question/index?qid=20100109211024AAoajAM
Sat, 09 Jan 2010 21:37:49 +0000
V(0)=rw ,w=2*pi*f, w=23.248,v(0)=0.52*23.248=12.089
so now we are dealing a projectile motion and we need too write protection path equation.
which will reads as: y=1/2*g*(x/Ettacos(teta ) )^2 + v(0)*sin(tata)*(x/v*cos(tata))+y(0)
putting data as:
y(0)=1.2 ,tata=38 degrees,y=2.4,v(0)=0.52 you will find x,which is your question.