Physicsplease checkthis can't be right
posted by Anthony .
I asked about a golf ball hit at a 35 degree angle and the hole was 175 m away. What was speed of ball. I got a response, and it follows with my response. I'm not doing something right, can you help, please.Physics  drwls, Saturday, October 9, 2010 at 10:20am
You can come up with an answer that would cause the ball to reach the hole on the fly, but in most cases you would want to allow for the ball to roll up to the cup, for a shot that long.
To reach the hole as distance X on the fly with an A = 35 degree launch angle, use
X = (V^2/g)*sin(2A) = V^2/g * 0.9397
Plug in X = 175 m and solve for V
PhysicsPlease check  Anthony, Saturday, October 9, 2010 at 11:23am
I don't think I'm doing this correct.
I took 175 = v^2/9.8 x 0.9397
I did 9.8 x 0.9397 = 9.02
then I took 175/9.20=19.02
This can't be rightdid I make a mistake with the variables? I'm not getting this Please help

YOu didn't do it right.
I took 175 = v^2/9.8 x 0.9397 right
which is 175*9.8/.9397=V^2 
First,is that the correct formula? I redid it, I think it should be 43 for the speed. Is that correct? If not, please direct me on what I'm doing wrong.

It is the correct relation. You need to start doing some analysis, not searching for "correct" formulas.
Consider the time in air: vertical
hf=hi+VsinThetat4.9t^2
VsinTheta=4.9t
t= you do it.
Then,horizontal
distance=VcosTheta*t above, and reduce it.
Remember that sin2A=2sinAcosA 
Thank you,
Sorry to have not understood itI apologize