Help Finding the initial speed as a diver leaves a board.

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SUMMARY

The discussion focuses on calculating the initial speed of a diver leaving a board 3.05 meters above water, with a final speed of 8.51 m/s at an angle of 75.1 degrees upon entering the water. The primary equation used is (Vf^2) = (Vo^2) + 2a(x), where Vf is the final velocity, Vo is the initial velocity, a is the acceleration (−9.8 m/s²), and x is the vertical displacement (−3.05 m). The correct initial vertical speed (Vo) calculated is 2.79 m/s, confirming the method's validity while addressing common calculation errors.

PREREQUISITES
  • Understanding of kinematic equations, specifically (Vf^2) = (Vo^2) + 2a(x)
  • Basic knowledge of trigonometry, including sine, cosine, and the Pythagorean theorem
  • Familiarity with the concept of acceleration due to gravity (−9.8 m/s²)
  • Ability to manipulate and rearrange equations to solve for unknown variables
NEXT STEPS
  • Study the derivation and application of kinematic equations in vertical motion scenarios
  • Learn how to apply trigonometric functions to resolve components of velocity
  • Explore the concept of instantaneous velocity and its implications in physics problems
  • Practice additional problems involving projectile motion to reinforce understanding
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Students studying physics, particularly those focusing on kinematics and projectile motion, as well as educators seeking to clarify concepts related to vertical motion and initial velocity calculations.

DeltaIceman
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Homework Statement


A diver springs upward from a board that is 3.05 m above the water. At the instant she contacts the water her speed is 8.51 m/s and her body makes an angle of 75.1 degrees with respect to the horizontal surface of the water. Determine her initial speed as she leaves the board.


Homework Equations


The equations I have to use for this problem were
(Vf^2)=(Vo^2)+2a(x)
x=1/2(Vo+Vf)t
x=(Vo)(t)+1/2(a)(t^2)
Vx=Vo+at
******Note all these equations are interchangable and can be used to find the y components also********

Basic trig functions are also allowed to be used such as cos=x/r and rearranging it to find the values you need. These trig functions also include the pathagorian theroem (sp?)

The Attempt at a Solution


scan0001.jpg


This is all my work that I've done for the equation or at least my last attempt. I tried it a couple different ways but this is the only one I felt like I was getting anywhere. Also The equation I used to get Vo or initial velocity was (Vf^2)=(Vo^2)+2a(x).


I'm sure I'm making a dumb mistake. I went through the rest of the homework problems fairly easy. Then I got to this one and was dumbfounded and decided to seek help elsewhere. I wasn't exactly sure if I could assume the height was 3.05m but I started thinking about the problem and since it states that we want the initial speed as she leaves the board. We could consider that velocity to be almost instantaneous and since the elapsed time is so small that the height was basically 3.05 m. But then again its magnitude is going upward and I wasn't exactly sure how that would through off the height. Anyway If you guys can help me it would be greatly appreciated. I am sorry for any inconvience or any time consumed to help me with this problem. Thank you very much.

~Eric
 
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Using your equation v_{fy}^{2}=v_{0y}^{2}+2ay with vf=8.22, a=-9.8 and y=-3.05 yields v0y=2.79m/s. I think you plugged something into your calculator incorrectly because your method is correct (not sure why you got a time value, as it is unneeded).
 
Thank you so much. I thought I made a really dumb mistake.
 

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