Golf ball problem (2D Kinematics)

[chudd88]

Homework Statement

A golf ball is chipped with an initial velocity of 20 m/s along a level fairway.

a) WHat angle should the initial velocity make with the horizontal for the maximum height to be equal to the horizontal distance on the fly?

b) What is this horizontal distance.

Homework Equations

The Attempt at a Solution

I've been working on this problem over the past couple of days, and I'm stuck on it. It's one of the problems at the end of the chapter, which are supposed to be the most challenging, but I feel like either there's some important mathematical formula I'm expected to remember, or there's something very simple I'm overlooking.

My attempts so far:

* The distance x at time t is the same as the height y at time t/2

* The initial horizontal and vertical velocities are 20*cos(theta) and 20*sin(theta) respectively.

* I've done every manner of substitution I can think of, converting values into corresponding values using all of the equations I have available to me. Either I come out with a totally wrong answer, or I just end up with an identity.


I understand the basics of 2D kinematics. If I knew the initial angle I'd be able to figure out everything about the ball's flight. But, trying to use what's available to me, and work it backwards to the angle, isn't working out for me. I keep wanting to get a value relative to cos(theta) or sin(theta), and use the pythagorean theorem to get one of the other initial velocities, but I'm just not getting there.

Could someone give me a hint as to what it's going to take to figure this out? I don't expect anyone to do the whole thing, but if you can see, looking at the problem, some bit of insight I'm obviously missing, that would be very helpful.

Thanks.

 

[LowlyPion]

Homework Statement

A golf ball is chipped with an initial velocity of 20 m/s along a level fairway.

a) WHat angle should the initial velocity make with the horizontal for the maximum height to be equal to the horizontal distance on the fly?

b) What is this horizontal distance.

I've been working on this problem over the past couple of days, and I'm stuck on it. It's one of the problems at the end of the chapter, which are supposed to be the most challenging, but I feel like either there's some important mathematical formula I'm expected to remember, or there's something very simple I'm overlooking.

My attempts so far:

* The distance x at time t is the same as the height y at time t/2

Try this:

t * V_x = x or V_x = X/T where T is the total time and X is total Distance which is also height.

We also know that X = 1/2*g* (T/2)²
(T/2 because what goes down must have gone up.)

Substituting then we know V_x = (1/2 * g * T²/4) / T = 1/8 * g * T

Now you also know that V_y = g * T/2

The ratio of V_y/V_x = Tanθ

 

[chudd88]

Thank you for the help. I need to go over your work a bit more to understand it, but I can see it leading me to the answer. The greatest help was in your statement:

The ratio of Vy/Vx = Tanθ

I realize that should have been obvious to me, but I just wasn't seeing it. That was the "insight" I needed to relate all the pieces together.

Thank you again.

 

[LowlyPion]

Thank you for the help. I need to go over your work a bit more to understand it, but I can see it leading me to the answer. The greatest help was in your statement:

The ratio of Vy/Vx = Tanθ

I realize that should have been obvious to me, but I just wasn't seeing it. That was the "insight" I needed to relate all the pieces together.

Thank you again.

Happy to help.

Cheers.