# Return on the Baffling Simple Physics Question

Once again, I find myself in need of extra help in the physics department.

If a golf ball goes 240m (delta x) when hit at an angle of 34.0[sup]o[/sup], what’s the initial velocity (V[sub]i[/sub]) and maximum height (delta y for half way through the trajectory) of the trajectory, with no air resistance?

I’ve looked at all the equations I know in the short time I’ve studied physics, and I can’t get any of them down to one unknown in order to solve for V[sub]i[/sub]. What is the concept I am missing to bust this open?

Perhaps what you are missing is the time link. The time it takes for the golfball to travel 240m in the x direction is the same as the time to travel up to its maximum height and back down to the earth.

I know that. I don’t know how to solve for time though (its not given,) as I said before, I can’t get any equation down to one unknown in order to solve for it.

I think Cheesesteak is right. I just did the problem by writing out the equations for v[sub]x[/sub] and v[sub]y[/sub], then eliminated v[sub]y[/sub] using the equation v[sub]y[/sub] = v[sub]x[/sub]*tan 34

I should have said: writing out the equations for v[sub]x[/sub] and v[sub]y[/sub] with respect to time.

delta x = (v[sub]i[/sub]cos(34))*t

delta y = (v[sub]i[/sub] sin(34))*t - (1/2)gt[sup]2[/sup]

Using these two equations ought to do it for you. I feel I should leave something for you to do but if you can’t figure it out let me know.

What is delta y.

Ring, forgive me for being thick skulled, but I don’t see how you can solve for Vi without knowing t. I considered those equations when I was pondering the problem, but dismissed them for that reason.

Also, on the second equation, shouldn’t it be delta y = (vi sin(34))*t + (1/2)gt[sup]2[/sup] ? or are you just calling it that because you’re saying g is positive?

Ring should have spelled it out…

delta x in his equation equals 240. From that given you can find time in terms of v sub i. Then you know that the delta y is zero. That gives you a nice quadratic to solve (two solutions… one at zero… duh! one at your answer for time). From there, it’s cake to use 1/2 the time to figure out using the y-component of your velocity how high the bugger gets in the air.

Hope you got it now.

Assumptions:
at t = 0, x[sub]0[/sub] = 0 and y[sub]0[/sub] = 0
a[sub]x[/sub] = 0
a[sub]y[/sub] = -9.8 m/s[sup]2[/sup]

Given:
x = 240 m
y = 0 (vertical displacement is zero over the whole trip)
v[sub]x0[/sub] = v[sub]x[/sub] = v[sub]0[/sub]cos34
v[sub]y0[/sub] = v[sub]0[/sub]sin34

Use:
x = v[sub]0/subt
y = v[sub]0/subt + 0.5a[sub]y[/sub]t[sup]2[/sup]

You have here two equations and two unknowns. Solve for v[sub]0[/sub] and t. Once this is done, the maximum height may be easily found–this is left as an exercise for the interested reader.

(I currently teach physics–you’re not getting the answer out of me, either! :D)

I don’t like to use g in equations, as students generally get the sign wrong more often than not. For the record, g is a vector that points downward with magnitude 9.8 m/s[sup]2[/sup]. Thus, the magnitude of g is a positive number.

It appears your questions from above have been answered, but just for your edification if you eliminate t in the equations I posted you wind up with a nice equation for the range. (You also have to complete the square)

R = (v[sub]i[/sub])[sup]2[/sup]/g]sin(2theta[sub]i[/sub])

This question has been kind of bothering me. Let me see if I can, in plain language, explain why it has to be negative.

You see (v[sub]i[/sub] sin(34)) is the vertical component of the initial velocity. This means that unless something counteracts this upward velocity the projectile will just keep going up forever.

However we both know this won’t happen because gravity will eventually pull it back down. So therefore in order to make the equation reflect reality we have to subtract the effect of gravity from the initial velocity.

Try to visualize what is happening. Equations aren’t magic formulas they have to reflect reality.

Ok, I finally cracked this bad boy open, around 9:30 (only 3 and a half hours spent on one question!) last night. I was trying to solve for everything right a way, instead of solving for t in terms of the other things, then plug that equation into the RBF (really big formula), as my phsyics teacher calls it) to get the answer.

I’d like to just say now that I wasn’t posting this to get the answer. I knew the answers (they’re in the back of the book), but didn’t know how to solve it. Neither did 95% of the people in my honors physics class.

Anyway, thanks for the help everyone.

And WRT g, my teacher told us g = -9.81m/s[sup]2[/sup]. Using that, the second part of the RBF will be negative (unless time is negative). Thus, by adding the first part to a negative number, you get the same answer as if you said g = +9.81 and just subtracted that.

As an interesting sidenote, my physics teacher was a former defense contracter who helped to build the guidance systems we’re now pounding Afghanistan with.

Followup to my post above regarding a[sub]y[/sub] and g:

Note that the magnitude of the vector g is 9.8 m/s[sup]2[/sup].

If we choose “upward” (away from the center of the Earth) to be the direction of the +y-axis, then:

a[sub]y[/sub] = -9.8 m/s[sup]2[/sup] = -g

and

g = 9.8 m/s[sup]2[/sup]

Also, note that if g was a negative number, than in the equation

w = mg

You would get a negative number for an object’s weight. This is unconventional, to say the least.

You don’t have to take my word for it. Out of the commonly used physics textbooks, including Tipler, Serway, Giancoli, and Halliday & Resnick, all use g = 9.8 m/s[sup]2[/sup]. The latter text explicitly states to not use a negative value for g.

I just thought that someone ought to point out that neglecting air resistance for golf balls is a bad approximation of the real world[sup]TM[/sup].
In fact, if it wasn’t for the air resistance (specificaly the Magnus Effect) the ball would travel shorter!

I’ll give here the advice I gave all students (and tutees) regarding physics problem solving.

1. Draw a picture. I don’t care if it a body at rest decomposing. Draw a picture.

2. Draw and label all forces (in this case, gravity should be drawn).

3. Choose a frame of reference. Interestingly enough, you can choose any direction for your x and y axes. You could make them 36.5 degrees off horizontal if you wish. It helps, however, to choose a frame that makes the math easier. In this problem, there are two obvious options:
a)34.0 deg from horizontal
b)0 deg from horizontal

I would have chosen 0 deg from horizontal. To choose the other would require decomposition of the gravity vector. With this frame I have to decompost the initial velocity vector.

1. Write down ALL information, known and requested.

2. Write down ALL standard form equations related to the motion (2D in this example).

Now you can choose equations and solve.

Students have a tendency to skip step 1 or 2 most often. This leads to problems. With practice and comfort with a particular problem type (rotational dynamics, kinematics, etc.) you might be able to skip steps 4 or 5.

For fun, qwertyasdfg, try this problem again using these steps. DON’T SKIP ANY!! See if it’s easier.

Best of luck and remember, Physics is Phun!

(if you need help, feel free to e-mail me; once a teacher…)

Spritle