Calculating Diver Velocity and Maximum Height in a 3.0m Board Dive

AI Thread Summary
The discussion focuses on calculating the diver's velocity upon striking the water and determining the maximum height reached above the water after jumping from a 3.0 m board with an initial speed of 1.7 m/s. The initial calculations for velocity were incorrect, particularly in the application of the formula v = d/t, which does not yield the correct final velocity. Participants suggest using energy conservation principles and kinematic equations, specifically those involving the squares of velocities, to solve the problem accurately. The correct approach involves considering both initial velocity and acceleration due to gravity. Understanding these concepts is crucial for solving both parts of the question effectively.
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A diver springs upward with an initial speed of 1.7 m/s from a 3.0 m board.

a. Find the velocity with which he strikes the water. (Hint: When the diver reaches the water, his displacement is y = -3.0 m (measured from the board), assuming that the downward direction is chosen as the negative direction.)

b. What is the highest point he reaches above the water?


I tried using this equation, but my answer was wrong. Can someone tell me what I did wrong?

Part a:
d=v*t + 1/2 at^2
-3.0=1.7t + 1/2(-9.80)t^2
4.9t^2 - 1.7t -3.0=0
t= 0.98s

v=d/t
v=3/0.98
v=3.06m/s

I didn't do Part b because Part a was wrong!
 
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When I did this question, I first noticed to use energy. For me, that's more comfortable.
However, your method also is a good method, even better than mine.
t is correctly found.

"v=d/t
v=3/0.98
v=3.06m/s" is wrong.
Try to find out what's the exactly meaning of the v in v=d/t.
You are going to find a particular v but not that one.
 
For the second question:HINT:Use Galiei's formula,the one with the squares of velocities.

Daniel.
 
A) Hint: Remember the initial velocity v(0):

: \ \ \ \ \ v(t) = v(0) + at

B) Formula with "velocities squared" is equivalent to Energy approach for constant acceleration "a":

: \ \ \ \ \ v^2(t) - v^2(0) = 2a(d(t) - d(0))

: \ \ \ \ \ \frac {mv^2(t)} {2} - \frac {mv^2(0)} {2} = (ma)(d(t) - d(0))

:: \ \ \ \ \ \Delta K.E. = (Force)(\Delta Distance)


~~
 
Double posting is not allowed...

Daniel.
 
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