First we must break it up into its individual x and y directions.
velocity in y direction =9.01*cos(11)= 8.8445 m/s downwards
velocity in x direciton =9.01*cos(79)=
1.71918 m/s to the right.
Given Vfy(velocity final y direction) we can find max height by using V=Vi+At acceleration being -9.81m/s^2.
using that formula we get 3.9869 meters as the max height reached.
As the diver starts 3 meters above the water 3.9869m-3m=.9869 m we can use this to find initial velocity in y direction using equation (vf)^2-(vi)^2=2a(xf-xi)
plugging this in given vfy = 0 at the top of the curve we get Viy= 4.4003 m/s
Because there is no horizontal acceleration Vix=1.71918 m/s (found at beginning). To get the magnitude of this velocity we use the Pythagorean thrm. 4.4003^2+1.7198^2=(4.724)^2
initial magnitude=4.724 m/s
Finally to determine direction we use inverse cosine of Vx/Vmagnitude to get a final answer of 68 degrees with respect to the horizontal.
equations used:
v²-v₀²=2a(x-x₀) v = v₀ + at
a^2+b^2=c^2 and inverse cosine
Olympic diver Matthew Mitcham springs upward from a diving board that is 3.00 m above the water. He enters the water at a 79.0 degree angle with respect to the water surface, at a speed of 9.01 m/s. Determine the magnitude of his initial velocity. Determine the direction of his initial velocity, in terms of degrees relative to horizontal. Determine his maximum height above the water.
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