@Study23
Sooo close! All of the concepts are there, just make sure that you realize in the equation
\[ v_f^2=2aH\] (which, again, I wrote while using variables willy nilly)
**the only velocity that contributes to H, the maximum height the diver gets to, is the Y velocity!** so it should be
\[v_{f}^2 = v_o^2 + 2ad \]
\[v_{fy}^2 = \cancel{v_{oy}^2} + 2a(H) \]
\[v_{fy}^2 = 2a(H) \]
So H is
\[ H =\frac{v_{fy}^2}{2a}=\frac{{(8.9m/s\sin75)}^2}{(2)(-9.81m/s^2)}=-3.77m\]
This equation says, "Starting from rest, a body must travel -3.77m - it must fall 3.77m meters below an initial location y=0 to attain a velocity of (8.9 sin 75)m/s given that acceleration is -9.81 m/s^2.
A very wordy confusing say of looking at it!... which we can get around by saying everything is a magnitude. Then the equation reads as "Starting from rest, a body under acceleration 9.81 m/s^2 has a velocity of (8.9 sin 75)m/s in the direction of acceleration after it has displaced 3.77m in the direction of acceleration" Then all of the positive and negative information is retrieved from defining your coordinate system on the picture.
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We can now say that the max height above the diving board , little h, is
\[ h = H - 3 = .77m\]
With that we're looking for the velocity of the diver after displacing .77 m
\[ v_{y@h}=\sqrt{2ah} = \sqrt{(2)(9.81m/s^2)(.77m)} = 3.76 m/s\]
"Starting from rest, after a body being accelerated at 9.81m/s^2 displaces .77m in the direction of the acceleration, it will be traveling a velocity of 3.76m/s in that direction."
Therefor, starting at the top of the trajectory and falling .77m (going .77m in the -y direction), its velocity will be 3.76 m/s in the negative y direction.
THIS means that at the other side of the parabola, its velocity will be equal in magnitude and opposite in direction! It's initial velocity in the y direction is 3.76 m/s in the positive y direction. Now we have v_xi and v_iy
\[ v_{xi} = v_{xf} = 8.9 \cos 75 m/s\]
\[v_{yi} = v_{y@h} = 3.76 m/s\]
Using these values to find our angle phi
\[ tan \phi = \frac{v_{yi}}{v_{xi}} = \frac{3.76m/s}{8.9 \cos 75 m/s}=1.59 \]
\[ \phi = arctan 1.56 = 57.34º\]
Now to find the magnitude of v_i, which is the square root of the sum of the squares of its components.
\[ |\textbf v_i| = \sqrt{|v_{xi}|^2+|v_{yi}|^2} = \sqrt {(8.9\cos75m/s)^2+(3.76 m/s)^2} = 4.4m/s \]
SO! ::breaths::
v_i = 4.4m/s @ 57.34º above the horizontal.
This makes physical sense. The angle entering the water should be steeper than the initial angle because of an increase in the magnitude of the v velocity due to gravity.