Max height given angle and distance

Click For Summary
SUMMARY

The discussion revolves around calculating the maximum height of a golf ball hit at an angle of 25.0 degrees, covering a horizontal distance of 301.5 meters. The initial approach involved using trigonometric relationships, specifically the tangent function, but led to an incorrect conclusion of 70.30 meters instead of the correct maximum height of 35.1 meters. The participant ultimately discovered the range formula, which states that the horizontal distance (X) can be calculated using the initial velocity, the sine of double the angle (2a), and gravitational acceleration (g). This formula provides a pathway to derive the necessary initial velocity for further calculations.

PREREQUISITES
  • Understanding of projectile motion principles
  • Familiarity with trigonometric functions, particularly sine and tangent
  • Knowledge of kinematic equations related to vertical motion
  • Basic grasp of gravitational acceleration (g = 9.81 m/s²)
NEXT STEPS
  • Study the derivation and application of the range formula for projectile motion
  • Learn how to calculate maximum height using initial velocity and angle
  • Explore kinematic equations for vertical motion in detail
  • Investigate the effects of varying launch angles on projectile trajectories
USEFUL FOR

Students in physics or engineering, educators teaching projectile motion concepts, and anyone interested in solving real-world problems involving angles and distances in sports or engineering contexts.

agentnan
Messages
9
Reaction score
0
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.
 
Physics news on Phys.org
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!
 

Similar threads

Replies
5
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 19 ·
Replies
19
Views
3K
  • · Replies 15 ·
Replies
15
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
Replies
43
Views
5K
  • · Replies 14 ·
Replies
14
Views
4K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K