Have I done this correctly? The instructor said the answer is a range of values.

[AndrewHelmerich]

 

[Deleted member 4993]

View attachment 14382

For d = 120, we get (using g = 32.)

2θ = 47.79 or 132.21

θ = 23.9^o and 66.1^o

and

For d = 160, we get (using g = 32.)

2θ = 80.99 or 99.01

θ = 40.5^o and 49.5^o

Thus to stick the ball to green:

23.9^o < = θ < = 66.1^o

 

[Cubist]

Actually there's a gap in the middle where the ball goes too far, see the attached graph of angle(x) vs distance(y)



So the actual answer has two ranges, highlighted in red, which I'm sure you can work out from the graph and the accurate numbers in Subhotosh Khan's post

 

[Deleted member 4993]

Actually there's a gap in the middle where the ball goes too far, see the attached graph of angle(x) vs distance(y)

View attachment 14391

So the actual answer has two ranges, highlighted in red, which I'm sure you can work out from the graph and the accurate numbers in Subhotosh Khan's post

Ooops.... I missed that! I had told myself that at ~45^o - there might be a hole. but at the end I forgot about it!! To the corner for 45 minutes.....

 

[Steven G]

Ooops.... I missed that! I had told myself that at ~45^o - there might be a hole. but at the end I forgot about it!! To the corner for 45 minutes.....

Again your in the corner? Is the academy going down hill from here?

 

[firemath]

View attachment 14382

View attachment 14382

That guy's pants are hilarious.