Max height given angle and distance

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A golfer is trying to determine the maximum height of a golf ball hit at a 25-degree angle over a horizontal distance of 301.5 meters. Initially, there is confusion about needing time or initial velocity to calculate the height, and the assumption that maximum height occurs at half the distance is proven incorrect. The golfer realizes that using the range formula, which incorporates initial velocity and angle, can help solve the problem. After finding the appropriate formula, the golfer expresses gratitude for the guidance received. The discussion highlights the importance of understanding projectile motion and the correct application of physics formulas.
agentnan
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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.
 
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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!
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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