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

[shawonna23]

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!

 

[primarygun]

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.

 

[dextercioby]

For the second questionINT:Use Galiei's formula,the one with the squares of velocities.

Daniel.

 

[xanthym]

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

[tex]: \ \ \ \ \ v(t) = v(0) + at [/tex]

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

[tex]: \ \ \ \ \ v^2(t) - v^2(0) = 2a(d(t) - d(0)) [/tex]

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

[tex] :: \ \ \ \ \ \Delta K.E. = (Force)(\Delta Distance) [/tex]


~~

 

[dextercioby]

Double posting is not allowed...

Daniel.