How Do You Calculate Initial Velocity in 2D Projectile Motion?

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SUMMARY

The discussion focuses on calculating the initial velocity of a diver in 2D projectile motion, specifically from a height of 3 meters with a final speed of 8.90 m/s at a 75-degree angle upon water contact. The diver's initial velocity components were analyzed using kinematic equations, particularly V2^2 = V1^2 + 2ad. The confusion arose regarding the choice of coordinate system, particularly the designation of positive and negative directions, which impacted the calculation of the initial vertical velocity.

PREREQUISITES
  • Understanding of 2D projectile motion principles
  • Familiarity with kinematic equations, specifically V2^2 = V1^2 + 2ad
  • Knowledge of trigonometric functions (sine and cosine) for angle calculations
  • Ability to apply vector resolution in physics problems
NEXT STEPS
  • Study the derivation and application of kinematic equations in projectile motion
  • Learn about vector resolution and its importance in physics problems
  • Explore the impact of coordinate system choices on motion analysis
  • Practice solving various projectile motion problems with different initial conditions
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 vector analysis.

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Homework Statement



A diver springs upward from a board that is 3m above the water. At the instant she contacts the water her speed is 8.90 m/s and her body makes an angle of 75 degrees with respect to the horizontal surface of the water. Determine her initial velocity, both magnitude and direction.

Initial velocity = V1 , Final Velocity = V2

Assuming Down as positive and up as negative
Y direction X direction
Dy = 3m. Dx = Unknown.
Ay = 9.8 Ax = 0
v2y = Sin 75 8.90 v2x = Cos 75 8.90
v1 = unknown v1x = v2x ( since there is no accelaration ! )



Homework Equations


V2^2=V1^2 + 2ad



The Attempt at a Solution



Since V1x = v2x , We do not need to calculate V1x since its already 2.3 ( cos 75 8.90 )

V2Y = ( Sin 75 8.90 ) = 8.60

V2y^2= V1y^2 + 2ad
(8.60) = V1Y^2 + (2x9.8x3)
(8.60)^2-(2x9.8x3) = V1y^2
3.89 = V1y^2

Since 3.89 is positive, it should be pointing down. But when i make a vector with y point down and x pointing to the right, after using pythagoras theorem and tan , i get the wrong degree. However, if i assume down to be as negative and up to be as positive, then my direction is correct.
I have always been told by my teachers that its upto me what i take as negative and positive. I want to know why I am not being able to get the answer when i take down as positive and up as negative. either way i get my Y value as positive whether i take positive down or up , so how do i know which one is right.

Help will really be appreciated. Thanks.
 
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