Projectile Motion: How to Calculate Initial Velocity for a Golf Shot

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SUMMARY

The discussion focuses on calculating the initial velocity required for a golf shot that travels a horizontal distance of 130 meters and reaches a vertical height of 20 meters before descending 10 meters below the tee-off point. The established initial velocity is 35.1 m/s at an angle of 34.4 degrees. Key equations used include the kinematic equations for projectile motion: S = V₀t + 1/2At², V = V₀ + At, and V² = V₀² + 2AS. The user successfully calculated the vertical component of initial velocity (V₀y = 19.8 m/s) and the time of flight (T = 2s) but struggled with the horizontal component.

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A golfer wants to hit the ball a horizontal distance of 130M and 20M above his tee-off point, but wants the ball to land 10 M below his tee-off point. What is the balls Initial velocity?


S= volt+1/2AT²
V= Vo+AT
V²=Vo²+2AS

So far I have broken up the graph into 3 Parts...from Point A to Point B i have solved for the Y direction.
Ay= -9.81m/s²
Voy= 19.8m/s
Vx= 0
Sy= 20m
T= 2s

But I can't seem to start off in the X direction...do I go from Point A to C? B to C in the Y direction? I am really lost everytime I try to solve another direction from ethier point I have no information to create a solution.

btw the answer is Vo= 35.1 m/s angled at 34.4 so please just help me from understanding how to go from here.
 

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You must do things in this order: determine the Y component of initial velocity, determine the time for the ball to reach C, determine the X component of the initial velocity using the time.

What part of these three steps are you struggling with?
 

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