Parabolic Projectile Motion problem.

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SUMMARY

The discussion focuses on solving a parabolic projectile motion problem involving a golf ball hit at a speed of 14.0 m/s at a 40-degree angle, aimed at a green elevated 3.00 m. Key variables include initial and final positions, initial velocity, and gravitational acceleration of -9.8 m/s². The participant, Cory, calculated the initial vertical and horizontal velocities as approximately 8.999 m/s and 10.725 m/s, respectively, and sought guidance on applying the relevant kinematic equations to determine the ball's speed just before landing.

PREREQUISITES
  • Understanding of kinematic equations for projectile motion
  • Basic knowledge of trigonometric functions (sine and cosine)
  • Familiarity with Newton's laws of motion
  • Concept of gravitational acceleration (9.8 m/s²)
NEXT STEPS
  • Calculate the time of flight using the vertical motion equations
  • Determine the horizontal distance traveled using the horizontal velocity
  • Learn how to apply the conservation of energy in projectile motion
  • Explore the effects of air resistance on projectile motion
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and projectile motion, as well as educators seeking to clarify concepts related to kinematics and forces in motion.

ssjcory
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Homework Statement


A golfer hits a shot to a green that is elevated 3.00 m above the point where the ball is struck. The ball leaves the club at a speed of 14.0 m/s at an angle of 40 degrees above the horizontal. It rises to it's maximum height and then falls down to the green. Ignoring air resistance, find the speed of the ball just before it lands.

So I say that we have these variables:
Yi(Initial y position) = 0
Yf(Final y position) = 3m
Xi(Initial x position) = 0
Xf(Final x position) = unknown
Vi(Initial velocity) = 14
Ti(Initial time) = 0
AYc(Gravity y accel) = -9.8
Initial angle of motion = 40 deg

Homework Equations


I think these equations are relevant
Horizontal Velocity = magnitude * Cosine(theta)
Vertical Velocity = magnitude * sin (theta)
Xf = Xi + VXi(Tf - Ti) + 1/2 Ax (Tf - Ti)^2
Yf = Yi + VYi(Tf - Ti) + 1/2 Ax (Tf - Ti)^2


The Attempt at a Solution


I drew a crappy diagram to get the vision of the green being higher than the tee.
My XY axis cross at ground level.

First I tried getting the direction specific velocities based on the angle and the initial velocity
VYi = 14 sin(40) == 8.999
VXi = 14 cos(40 == 10.725

I guessed that I would substitute those into the 3rd/4th equations listed up top. But I didn't know Ax so I assume it is constant? I am so confused about where to go from here.

Can anyone point me in the right direction?

Thanks,
Cory
 
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The whole idea behind projectile motion is that it is motion of an object that is in free fall, meaning that it is under the influence of gravity only. No other forces are acting. As a result, Newton's 2nd law says that a_x = 0. There is no horizontal acceleration. That simplifies your x equation somewhat.
 

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