Max height given angle and distance

[agentnan]

Good morning,

I am trying to set up the following problem:

A golfer hits a golf ball at an angle of 25.0 degrees to the ground. If the golf ball covers a horizontal distance of 301.5 m, what is the ball's maximum height? (Hint: At the top of its flight, the ball's vertical velocity component will be zero).

I keep coming back to needing to know either time &/or the initial velocity.

Can it be assumed that the maximum height will occur at half the distance of X? If it can, I could use:
tan25 = max height of Y / divided by .5X
this would give me a solution for the max height of 70.30. Unfortunately this answer is wrong. According to the book the answer is 35.1m, so I gather I can't do this...

I then tried based on the hint to use the formula for the final velocity in the y direction as 0, but in both formulas which involve the final velocity in the y direction I would need to know the initial velocity which I do not have. To find the initial velocity I would need the time. In all the formulas involving initial velocity or time, I am missing at least 2 factors, so I do not see a way to solve for either of them.

I am sure I am missing some facet of this question. Any guidance would be appreciated.

 

[agentnan]

please disregard...I just found a range formula for X that was later in the chapter which I can use. (change in X = initial velocity X sin 2a / g ) . Thanks for your help in this matter!

 

[kyle8921]

Dear golfer,

Thank you for your question. Finding the maximum height of the golf ball can be tricky, but with the right approach, we can solve it. You are on the right track with using the formula for the final velocity in the y direction as 0. This is because at the top of the ball's flight, its vertical velocity will be zero, meaning it has reached its maximum height.

To solve for the initial velocity, we can use the formula for the horizontal distance covered by the ball, which is given as 301.5 m. The formula for horizontal distance is d = v0cosθt, where v0 is the initial velocity, θ is the angle, and t is the time. Since we know the angle and the distance, we can rearrange the formula to solve for the initial velocity: v0 = d/(cosθt).

Next, we need to find the time it takes for the ball to reach the maximum height. We can use the formula for vertical velocity, which is given as v = v0sinθ - gt, where v is the final velocity (which is 0 at the top of the flight), v0 is the initial velocity (which we just solved for), θ is the angle, and g is the acceleration due to gravity (9.8 m/s^2). We can rearrange this formula to solve for time: t = v0sinθ/g.

Now that we have the initial velocity and the time, we can use the formula for maximum height, which is given as h = v0sinθt - 1/2gt^2. Plugging in the values we have, we get h = (d/(cosθt))sinθt - 1/2gt^2. Simplifying this, we get h = d(tanθ) - 1/2gt^2. Plugging in the values for θ, d, and t, we get h = 301.5(tan25) - 1/2(9.8)(7.53)^2 = 35.1 m. This is the same answer given in the book, so we have solved the problem correctly.

I hope this explanation helps you understand the problem better. Keep in mind that in physics, there are often multiple ways to solve a problem, so if you have a different approach that works for you, that is also valid. Keep up