Solve Golf's Ocean Dilemma: Projectile Motion Help for a 20 Degree Drive

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Homework Help Overview

The problem involves projectile motion, specifically analyzing the trajectory of a golf ball hit from a height of 4.0 meters at an angle of 20 degrees. The ball lands in the ocean after 5.0 seconds, prompting questions about its initial velocity, maximum height, horizontal distance traveled, and the time taken for the sound of the splash to reach the golfer.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants express uncertainty about how to begin solving the problem. Some suggest using specific equations related to projectile motion, while others question the origin of these equations and their applicability. There are discussions about separating the motion into horizontal and vertical components and using known values to find the initial velocity.

Discussion Status

Participants are actively engaging with the problem, offering various equations and approaches to tackle the different aspects of the projectile motion. There is a mix of suggestions and clarifications being provided, but no consensus has been reached on a single method to solve the problem.

Contextual Notes

Some participants note the potential confusion between 2D and 3D motion, emphasizing that the problem can be approached as a 2D motion scenario. There are also references to gravitational acceleration and the need for further clarification on certain equations used in the discussion.

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A golfer on a hill 4.0m above sea level hits a drive at an angle of 20 degrees, but unfortunately slices it to the right so that it goes into the ocean! If the ball lands in the briny 5.0 seconds after it was hit, what is the initial velocity of the ball? What height (above the cliff) did the ball reach before falling downward into the water? How far from the tee did it travel? How much later will he hear the splash of the ball? Assume sound travels at 343 m/s.
 
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I am not even really sur ehow to get started with this problem!
 
There is an equation like
y = xtantheta - [gx^2/(2v^2cos^2theta)] where x is horizontal distance covered andy is vertical distance. Use this and report if you still have problem. Also try deriving the above formula.
you must also be knowing that t = 2vsintheta/g
 
Where did the equation involving the tangent come from? I do not remember my instructor going over any equations with a tangent in them.
 
First find t fromthe horixontal equation x = vcostheta*t.
then substitute t in the equation y = vsintheta*t - 1/2g*t^2
 
Juz letta me give u a little push in starting this qn..
It may seem like a 3D motion qn but it is actually juz 2D..

For the 1st part, u noe that the ball touches the water 5s after the golfer hits the ball.. u oso noe that the height (oso known as the displacement) above the water where the golfer hits is 4m.. u oso noe that the vertical acceleration for bodies in 2D motion is only gravity ( 9.81m/s^2) and what u wan now is the initial velocity.. note: <---this is vertical.. u still nid to get use pythagoras theorem on both the horizontal and vertical velocities to get the initial velocity..

I hope this gets u start off.. the rest shld be quite easy..
 

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