Golf ball problem (velocity, max height, hang time)

[aquaorange]

"golf ball problem" (velocity, max height, hang time)

Okay. I'm having a really hard time with this. I know how to start the problem, but after that I go nuts, and literally start pulling out my hair. There is no book I can look in for reference, as this part is teacher supplemented.

The given picture/information is this:

And I need to find:

velocity of x at zero time:
velocity of y at zero time:
maximum height (from ground, not cliff):
hang time:
velocity of x after "a" time:
velocity of y after "a" time:

Now, I can find Vx at zero (20m/s), and Vy at zero (34.6m/s). I also know that Vx at any time is going to remain the same (20m/s), but then I'm lost for maximum height, hang time, and Vy at "a" time. Gravity is taken into account, but not air resistance. I know that it loses -9.8m/s going up due to gravity, and +9.8m/s going down due to gravity. And I tried doing that a long way, subtracting 9.8 from 34.6, subtracting 9.8 form that answer, etc., and counting how many times to get a seconds number, but then I didn't know how to figure out when I got down to 5.2 and subtracting 9.8 from that. I really have no clue how far up it goes, or how to find anything past making the triangle to find the forces using sine/cosine/tangent.

I'm not sure if there is a formula or multiple formulas to figure this information out. That's what I think I'm missing to get the rest of the information.

 

[radou]

The formula which will be of great use to you is the displacement formula. For the x direction, you have $$x(t) = x_{0} + v_{0x}t$$, and, for the y direction, $$y(t) = y_{0} + v_{0y}t - \frac{1}{2}gt^2$$. You should be able to solve the problem now.

 

[kyle8921]

Hello, it seems like you are struggling with the golf ball problem. Don't worry, there are a few formulas that can help you solve this problem. First, let's start with the basics: the velocity of the golf ball at any given time can be calculated using the formula V=V0+at, where V0 is the initial velocity, a is the acceleration (in this case, due to gravity), and t is the time. As you mentioned, Vx remains constant at 20m/s, so we can use this formula to find the velocity of y at any time. For example, at time "a", we can use the formula Vy=Vy0+at, where Vy0 is the initial velocity in the y-direction (34.6m/s) and a is the acceleration due to gravity (-9.8m/s^2). This will give us the velocity of the ball at any given time.

To find the maximum height, we can use the formula h=Vy0^2/2a, where h is the maximum height, Vy0 is the initial velocity in the y-direction, and a is the acceleration due to gravity. This will give us the height at which the ball reaches before falling back down.

To find the hang time, we can use the formula t=2Vy0/a, where t is the hang time, Vy0 is the initial velocity in the y-direction, and a is the acceleration due to gravity. This will tell us how long the ball stays in the air before hitting the ground.

I hope these formulas will help you solve the problem. Remember to always double check your units and plug in the correct values. Good luck!