The Best Angle for Maximum Spitting Distance?

[IDumb]

I came across this problem in a story I was reading. I thought it would be easy... but I am having a great deal of trouble.

The story claims that physicist David Bohm was asked the question: if a person of height h stands on ground and holds their head at an angle theta relative to ground, then spits with an initial velocity of magnitude v₀. What angle should they hold their head at to maximize the distance the spit travels?

In the story, he was able to solve this problem in his head in seconds. (Who knows if the story is true/exaggerated). I can not get a reasonable answer in hours with pencil/paper, so I would appreciate some help before I bang my head into a wall.

I have proven that if h = 0 the optimal angle is 45 degrees. So would a nonzero h affect that? If so, how can i derive an equation to find the angle?

My solution:
I derived equations for the x and y components of position as functions of times, by integrating the known accelerations of 0 for x (neglecting air resistance) and -9.8 for y.

I set the y position function = 0 and solved for t using the quadratic equation. This yielded a positive and negative time... obviously I used the positive.

At this time, where y = 0, we know that x must = D, where D is the distance the spit travels before hitting the ground. So I subbed the time gotten from the quadratic equation into the x position function.

I subbed in V_x = v₀cos(theta) and v_y = v₀sin(theta) for the v_x's and v_y's in my equation for D.

I then took the derivative with respect to theta, set it = 0, and used a computer to solve for theta (using random numbers for v₀ and h).



So that was my solution. However, it is incorrect. When i varied the magnitude of the initial velocity, I received different answers, which I feel is incorrect (?). So is my method wrong, or am I just too stupid to do basic math >.<.

I would really REALLY appreciate some help on this... wasted hours of time on it. If you need me to clarify something or want to see my math I'll be happy to post it

Thanks

 

[berkeman]

I came across this problem in a story I was reading. I thought it would be easy... but I am having a great deal of trouble.

The story claims that physicist David Bohm was asked the question: if a person of height h stands on ground and holds their head at an angle theta relative to ground, then spits with an initial velocity of magnitude v₀. What angle should they hold their head at to maximize the distance the spit travels?

In the story, he was able to solve this problem in his head in seconds. (Who knows if the story is true/exaggerated). I can not get a reasonable answer in hours with pencil/paper, so I would appreciate some help before I bang my head into a wall.

I have proven that if h = 0 the optimal angle is 45 degrees. So would a nonzero h affect that? If so, how can i derive an equation to find the angle?

My solution:
I derived equations for the x and y components of position as functions of times, by integrating the known accelerations of 0 for x (neglecting air resistance) and -9.8 for y.

I set the y position function = 0 and solved for t using the quadratic equation. This yielded a positive and negative time... obviously I used the positive.

At this time, where y = 0, we know that x must = D, where D is the distance the spit travels before hitting the ground. So I subbed the time gotten from the quadratic equation into the x position function.

I subbed in V_x = v₀cos(theta) and v_y = v₀sin(theta) for the v_x's and v_y's in my equation for D.

I then took the derivative with respect to theta, set it = 0, and used a computer to solve for theta (using random numbers for v₀ and h).



So that was my solution. However, it is incorrect. When i varied the magnitude of the initial velocity, I received different answers, which I feel is incorrect (?). So is my method wrong, or am I just too stupid to do basic math >.<.

I would really REALLY appreciate some help on this... wasted hours of time on it. If you need me to clarify something or want to see my math I'll be happy to post it

Thanks

When I sketch a few parabolas, it does look like a lower angle gets you farther when you start from a non-zero height.

Yes, please post your work -- it sounds like you are on the right path basically (well, except for varying Vo). Putting in a non-zero Yo should cause the optimum angle to be lower than 45 degrees, I think.

 

[AlephZero]

So that was my solution. However, it is incorrect. When i varied the magnitude of the initial velocity, I received different answers, which I feel is incorrect (?). So is my method wrong, or am I just too stupid to do basic math

Your math was probably right even though you didn't believe it. The best angle DOES depend on the initial speed.

You can see this must be true without doing any math, if you think about the case where you start from below ground level rather than above it. Suppose the initial speed is just big enough to reach height h. The best you can do is spit vertically upwards and get a horizontal distance of 0. If you spit faster, you can aim at an angle to the vertical and get a horizontal distance bigger than 0. So it seems very plausible that the best angle depends on the speed.

When you do the math, the dimensionless quantity 2gh/v² comes into the equations, where h is the height, v is the intial velocity, and g is the acceleration of gravity.

You probably know what the equation v² = 2gh tells you about the motion of projectiles.

If h = 0, then the best angle (45 degrees) does not depend on v, but that is a special case because if h = 0 then 2gh/v² = 0 for any value of v.
But I'm not going to spoil your fun by doing all the math for you...