# Thread: hard mechanics problem :)

1. hey, sorry i couldnt get a picture so i will have to explain the diagram. (its quite simple)

its a projectile problem; imagine a golf course and you have to hit a golf ball out of the bunker with just enough speed to land in the hole. All you know is that the depth of the bunker is 2m and the horizontal distance to the hole is 15m. find the angle the ball has to be projected so that the ball lands in the hole. no wind or air resistance. no spin.

if there are any questions please just post. i have a solution so it can be done.

2.

3. It looks like a homework question, so explain how you think you need to answer it and I'll point out any errors or suggest a better method. It wouldn't be learning if I just did it for you

4. Is your goal to derive the angle of least velocity?

Might I suggest thinking about it in this way:

To drive the ball to the hole, the golfer will have to the hit the ball somewhere between two extremes. Either the golfer will have to hit the ball at the shallowest possible angle at a very high speed, or the golfer will have to hit the ball at the highest possible angle (excluding 90 degrees) at a very high speed. Between the two extremes, the required velocity for a given range will drop. More specifically, the required speed to get the ball in the hole will drop as you raise the angle from a near-zero degree until the velocity reaches a minimum required velocity somewhere in that range and begins to rise again as you raise the angle nearer to the high-angle extreme. You can derive it yourself from the projectile trajectory equation. It might involve some knowledge of calculus though.

Also, to test my working of the problem. Is the answer 48.797 degrees?

5. Originally Posted by rhysboi1991
hey, sorry i couldnt get a picture so i will have to explain the diagram. (its quite simple)

its a projectile problem; imagine a golf course and you have to hit a golf ball out of the bunker with just enough speed to land in the hole. All you know is that the depth of the bunker is 2m and the horizontal distance to the hole is 15m. find the angle the ball has to be projected so that the ball lands in the hole. no wind or air resistance. no spin.

if there are any questions please just post. i have a solution so it can be done.
Not enough information presented to give a definite solution.

6. I assume no rolling into the hole either.

7. sorry it should be the minimum possible angle, yes 48.8 degrees was correct and there is enough information to answer the question. no rolling.

it was a revision question which i had already done before posting.

i found it quite difficult because i had never seen a question like that and just liked to see how other people would get to that answer. I did it the 'longwinded way' not using calculus and using the quadratic formula 3 times (about 2 A4 pages of solution).

8. Originally Posted by rhysboi1991
sorry it should be the minimum possible angle, yes 48.8 degrees was correct and there is enough information to answer the question. no rolling.

it was a revision question which i had already done before posting.

i found it quite difficult because i had never seen a question like that and just liked to see how other people would get to that answer. I did it the 'longwinded way' not using calculus and using the quadratic formula 3 times (about 2 A4 pages of solution).
given the qualities of this latest qualifier, the minimum angle is not 48.8 degrees.

The proper qualifier you needed, which was not in your op, is what is the lowest velocity necessary. That was absent, so no, there was NOT enough information to give a definitive answer. Spo made an assumption and solved based on that, and it happened to be the correct one. As stated, however, it was an ambiguous question.

9. Ah, well if you already did it, I'll admit to the way I found it.

I started with the trajectory equation:

Plugging in a=b=0, y=2, x=15, g=9.81, rearranging the equation to "velocity equals", and using the trig. identity, tangent squared of the angle plus one = secant squared of the angle, I obtained:

From here, I used calculus to obtain the derivative of velocity with respect to angle:

The minimum velocity angle will occur when the above derivative equals 0. Equating the derivative to zero, using the reverse of the same trig identity as above and rearranging the equation will yield a quadratic equation:

Solving this equation yield two answers:

These are, of course, the numerical answers for the tangents of the angles. So, to find the angles, one must take the inverse tangent. Doing this for both answers will give you a possibility of four answers to choose from. Upon analysis, though, only one will make sense:

Arcane, why isn't there enough information? My only assumption was that the lowest velocity occurs where the derivative with respect to the angle equals zero. Also, the other three angles either point into the ground or in the opposite direction of the hole. The velocity ends up being 12.96 m/s for me. Why is the angle not correct?

10. Originally Posted by TheDr.Spo
Ah, well if you already did it, I'll admit to the way I found it.

I started with the trajectory equation:

Plugging in a=b=0, y=2, x=15, g=9.81, rearranging the equation to "velocity equals", and using the trig. identity, tangent squared of the angle plus one = secant squared of the angle, I obtained:

From here, I used calculus to obtain the derivative of velocity with respect to angle:

The minimum velocity angle will occur when the above derivative equals 0. Equating the derivative to zero, using the reverse of the same trig identity as above and rearranging the equation will yield a quadratic equation:

Solving this equation yield two answers:

These are, of course, the numerical answers for the tangents of the angles. So, to find the angles, one must take the inverse tangent. Doing this for both answers will give you a possibility of four answers to choose from. Upon analysis, though, only one will make sense:

Arcane, why isn't there enough information? My only assumption was that the lowest velocity occurs where the derivative with respect to the angle equals zero. Also, the other three angles either point into the ground or in the opposite direction of the hole. The velocity ends up being 12.96 m/s for me. Why is the angle not correct?
He didn't state he wanted the angle with lowest velocity, nor anything else about the velocity, other than that the ball would be hit with a velocity that put it in the hole.

11. Originally Posted by rhysboi1991
...you have to hit a golf ball out of the bunker with just enough speed to land in the hole.
I took "just enough speed" to mean the minimum velocity. Different verbiage, yes, but I believe the author will agree that this constituted the "minimum velocity" requirement.

12. yeah that's what I meant Dr.Spo, sorry for the confusion. i will put my solution up tomorrow its bed time where I am good night

13. when i said longwinded i meant i started with a basic suvat equation:

then re-arranged:

as it is a quadratic, i used quadratic formula to get t:

then i looked at the horizontal component and equalised both components in terms of t:

then i rearranged a little:

from here i looked for a trig identity involved tan^2 and sec^2

then substituted this identity and re-arranged (also divide through-out by g:

another quadratic but this time we say that the discriminant = 0, as we are looking for the minimum:

from here we use the other bit of the quadratic formula:

therefore:

which when rearranged gives another quadratic:

solved for tan(theta) gives:

Through analysis it only makes sense if it is the positive root:

sorry for the overuse of brackets, i had to use them to keep things in the right place. i didnt know the trajectory equation could of saved me alot of time. There may be a few little mistakes...its the first time i have used mathtype.

14. Your method of combining the x and y components of position brought you rather close to the trajectory equation I started with. Although, you didn't run with that and took an approach that didn't involve calculus, such as what I had done. Your way works and it produces the same correct result. One can plainly see that my quadratic equation has a = x, b = -2y, and c = -x. Although I went through the problem with the numbers on board instead of the symbols, the numbers are not mathematical coincidence. Your method shows that as well and explicitly so. This looks, as you said, long-winded. Do you have any knowledge of calculus? It might have been easier for you using it.

15. yeh im a 1st year physics student, but when i was doing the question i didnt think of using it at the time, then spent a while doing it that way

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