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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

Physics-Please 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 right-did I make a mistake with the variables? I'm not getting this- Please help

• Physics-please check-this can't be right - ,

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

• Physics-please check-this can't be right - ,

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.

• Physics-please check-this can't be right - ,

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+VsinThetat-4.9t^2
VsinTheta=4.9t
t= you do it.
Then,horizontal
distance=VcosTheta*t above, and reduce it.
Remember that sin2A=2sinAcosA

• Physics-please check-this can't be right - ,

Thank you,
Sorry to have not understood it-I apologize